romanian mathematical magazine r.m.m....anh, nguyen van mau, pham kim hung, vo quoc ba can, vasile...

Romanian Mathematical Society-Mehedinți Branch 2019

1 ROMANIAN MATHEMATICAL MAGAZINE NR. 22

ROMANIAN MATHEMATICAL SOCIETY

Mehedinți Branch

ROMANIAN MATHEMATICAL MAGAZINE

R.M.M.

Nr.22-2018



ROMANIAN MATHEMATICAL SOCIETY

Mehedinți Branch

DANIEL SITARU-ROMANIA EDITOR IN CHIEF ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT

ISSN 1584-4897 GHEORGHE CĂINICEANU-ROMANIA

EDITORIAL BOARD

DAN NĂNUȚI-ROMANIA EMILIA RĂDUCAN-ROMANIA MARIA UNGUREANU-ROMANIA DANA PAPONIU-ROMANIA GIMOIU IULIANA-ROMANIA DRAGA TĂTUCU MARIANA-ROMANIA CLAUDIA NĂNUȚI-ROMANIA DAN NEDEIANU-ROMANIA GABRIELA BONDOC-ROMANIA OVIDIU TICUȘI-ROMANIA

ROMANIAN MATHEMATICAL MAGAZINE-INTERACTIVE JOURNAL ISSN 2501-0099 WWW.SSMRMH.RO

DANIEL WISNIEWSKI-USA EDITORIAL BOARD VALMIR KRASNICI-KOSOVO

ALEXANDER BOGOMOLNY-USA



CONTENT

Algebraic inequalities from geometrical configurations- Daniel Sitaru,Claudia Nănuți........................................................................................................................................................................4

The beauty of inequalities-Do Huu Duc Thinh........................................................................................7

Elemente de geometria triunghiului in coordonate baricentrice-Neculai Stanciu...............20

Few applications of Cocea’s inequality- Daniel Sitaru,Claudia Nănuți...........................................25

About problem UP.147-ROMANIAN MATHEMATICAL MAGAZINE-Marin Chirciu..……………28 About problem 3677-CRUX MATHEMATICORUM- Marin Chirciu……. ……………………………….32

About problem JP.173-ROMANIAN MATHEMATICAL MAGAZINE- Marin Chirciu................35

About INEQUALITY IN TRIANGLE-884-ROMANIAN MATHEMATICAL MAGAZINE- Marin Chirciu.......................................................................................................................................................................37

Plynomes and matrices-Marian Ursărescu………………………………………………………………………39

Proposed problems………………………………………….………………………………………………………………...41

Index of proposers and solvers RMM-22 Paper Magazine.………………………………………….…………99



ALGEBRAIC INEQUALITIES FROM GEOMETRICAL CONFIGURATIONS

By Daniel Sitaru, Claudia Nănuți – Romania

Abstract: In this article we build geometrical configurations and extract algebraic inequalities from geometrical properties.

Aplication 1 : If , , ≥ 0 then:

1.1. + + √2 + + + √2 ≥ +

1.2 + + √ + + ≥ + + √2

1.3 + + √2 + + ≥ + + √2

Solution: Let be Δ such that: = ; = ; = ; = ;

= =34

By cosine theorem: = + ; = + + √2; = + + √2

In Δ : + > ; + > ; + >

+ + √2 + + + √2 ≥ +

+ + √2 + + ≥ + + √2

+ + √2 + + ≥ + + √2

Equality holds for = = = 0.



Aplication 2: If , , ≥ 0 then:

2.1. + + √ + + ≥ + + √3

2.2. + + + + √3 ≥ √ + +

2.3. √ + + + + + √3 ≥ +

Solution: Let be Δ such that: = ; = ; = ; = ; = ;

=56

By cosine theorem: = + ; = √ + + ; = + + √3

In Δ : + > ; + > ; + >

+ + + + ≥ + + √3

+ + + + √3 ≥ + +

+ + + + + √3 ≥ +

Equality holds for = = = 0.


3.1. + + √3 + + + √3 ≥ + +



3.2 + + √3 + + + ≥ + + √3

3.3. + + √3 + + + ≥ + + √3

Solution: Let Δ such that: = ; = ; =

= 60°; = =56

By cosine theorem: = + + ; = + + √3; = + + √3

In Δ : + > ; + > ; + >

+ + √3 + + + √3 ≥ + +

+ + √3 + + + ≥ + + √3

+ + √3 + + + ≥ + + √3


+ + + + + ≥ 2 3( + + )

Solution:



Let be a tetrahedron such that: ⊥ ⊥ ⊥ ; = ; = ; =

=+

; = + + ; =+ +

+

[ ] =⋅

2 =12 ⋅ + ⋅

+ ++

[ ] =+ +

2

By Mitrinovic’s inequality in Δ : ≥ √3 ⇒ ≥ √3 ⇒ ≥ √3

+ + + + √ +4 ≥

√32 ⋅ + +

+ + + + + ≥ 2 3( + + )

[1] Mihály Bencze; Daniel Sitaru – “699 Olympic Mathematical Challenges” – “Studis”

Publishing House – Iași – 2017

[2] Mihály Bencze; Daniel Sitaru – “Quantum Mathematical Power” – “Studis” Publishing

House – Iași – 2017

[3] Mihály Bencze; Daniel Sitaru – “Olympic Mathematical Energy” – “Studis” Publishing


[4] George Apostolopoulos; Daniel Sitaru – “The Olympic Mathematical Marathon” –

“Cartea Românească” Publishing House – Pitești – 2017

[5] Romanian Mathematical Magazine – Interactive Journal – www.ssmrmh.ro

THE BEAUTY OF INEQUALITIES

By Do Huu Duc Thinh-Vietnam

Inequality is one of the nicest and hardest major parts of Mathematics, because of its appearance in Math contests; and also it requires a solid knowledge to solve an inequality problem. But because of the elegance, Inequality attracts lots of student generations joining in solving and creating more. To maintain the passion and creativeness, many sites and forums nowadays have shared lots of topics and documents of inequality for everybody, but it’s still intricate and note arranged entiringly. Maybe many people will consider that studying Inequality is boring, since its large amount of knowledges and complex presentation of that amount of knowledges, especially upper secondary high school and



higher. But I will demonstrate carefully and clearly for everyone through this document file, so everyone can realize how beautiful inequality is and its application. This file that I make is based on documents of many teachers in Vietnam and around the world, such as: Tran Quoc Anh, Nguyen Van Mau, Pham Kim Hung, Vo Quoc Ba Can, Vasile Cîrtoaje,… I spent lots of time to compile the file so if there’s a mistake, I hope everybody can understand and give me feedback because my knowledge is limited. All feedbacks can be sent to the following email: [email protected] or Facebook. Thank you very much! Season 1: Some old and modern techniques of Inequalities in Math contests In this season, I will state again some inequalities and basic techniques that are useful to find the solution of proving the inequality and solving Min-Max problems. Also I will add more lemmas with inferences and developments so everyone, especially students, can find a nice way to solve the inequality. I. Inequalities can be proved by equivalence:

1) + ≥ ( ) ≥ 2 (Basic BCS inequality for 2 numbers)

→ ≥ ≥ √ ≥ = (Basic RMS-AM-GM-HM inequality for , > 0)

→ ≤ ≤ + − with , ≥ 0 such that + > 0

2) + + ≥ ( ) ≥ + +

→∗ 3( + + ) ≥ ( + + ) ≥ 3 ( + + )

∗ + + ≥ 3( + + ) ≥ + + ≥ 3( + + ) , , > 0

3) + + ≥ 2 + 2 − 2 4) + + ≥ 3 ∀ , , such that + + ≥ 0 (special case: , , > 0 – Cauchy’s inequality for 3 non-negative numbers) 5) ( + )( + ) ≥ ( + ) (Basic Bunyakovsky’s inequality) 6) + ≥ ( + )∀ , such that + ≥ 0 → 4( + ) ≥ ( + ) ≥ 4 ( + ) for + ≥ 0 7) 2( + − ) ≥ + ≥ 2 (2 − 3 + 2 )∀ ,

→ √ − + ≥ = + − with + ≠ 0

8) + ≥( )

with real numbers , such that > 0 (special case: , > 0)

9) + + ≥ with , , > 0 (Nesbitt’s inequality) → This is even true for real numbers ; ; such that + + > 0. 10) ( + )( + )( + ) ≥ ( + + )( + + ) ≥ 8 for , , ≥ 0.

→∗ ≥ ( + − )( + − )( + − ) , ,

∗ ( + + ) ≥ 27 , , ≥ 0( − )

11) + ≥ ( ) with , > 0 (Basic Bunyakovsky Cauchy-Schwarz inequality)

* Also, from this inequality we have the chain: + ≥ 2( + ) ≥ + ≥ 2√ for , > 0.

12) ∗ 2( + ) ≥ ( + )( + ) ≥ 2( + ) ℎ ≥ , ≥∗ 3( + + ) ≥ ( + + )( + + ) ≥ 3( + + ) ℎ ≥ ≥ , ≥ ≥ (Basic

Chebyshev’s inequality)



13) + + ≥ + + with ≥ ≥ * More general: + + ≥ + + for ≥ ≥ ≥ 0, ∈ 14) 2( + ) ≥ ( + )( + ) with , > 0; , , ∈ : + = → + ≥ + for , ≥ 0, , ∈

15) + + ≥ ( + ) ; − + ≥ ( + ) ∀ , → 3 ≥ ≥ with real numbers , : + > 0. 16) √ + + √ + ≥ ( + ) + ( + ) (Basic Minkovsky’s inequality) 17*)

( )^+

( )≥ with , ≥ 0 (The equality happens iff = = 1)

18) Consider ( ; ) = + − with , > 0. If ≥ 1 then ( ; ) ≥ 0; if ≤ 1 then ( ; ) ≤ 0.

19) + ≥ with , > 1 20) ( + + )( + + ) ≥ 3. max{ + + ; + + } with

, , ≥ 0 21) ( + + ) ≥ ( + + )( + + )∀ , , > 0 22) + + ≥ ≥ + + with ≥ ≥ > 0 23*)( + + ) ≥ 3( + + ) (Vasile’s inequality) 24) ≥ 1 − with ≥ 0 → + + ≥ ∀ , , ≥ 0: + + = 3. 25) ( + + ) ≥ + + + 24 ∀ , , > 0 II. Some familiar Inequalities, lemmas and techniques: (ascending by higher level) a) For junior – early – senior:

1) ⋯ ≥ ⋯ ≥ … ≥⋯

with > 0∀ = 1; 2; … ;

and ∈ , ≥ 2 (RMS-AM-GM-HM inequality for positive numbers) → If ; ; … ; are positive real numbers that + … + = 1 then with same condition for we have:

+ + ⋯+ ≥ ⋅ … (Weighted AM-GM inequality) 2) For real numbers ; ( = 1; 2; … ; ) we have: ( + + ⋯+ )( + + ⋯+ ) ≥ ( + + ⋯+ )

(Cauchy-Schwarz inequality) → + + ⋯+ ≥ ( ⋯ )⋯

∀ > 0

→ + + ⋯+ + ≥ ( + + ⋯+ ) ≥ + + ⋯+ ;∀ > 0

3) Let ≥ ≥ ⋯ ≥ ∗ ≥ ≥ ⋯ ≥ ℎ : ( + + ⋯+ ) ≥ ( + +⋯+ )( + + ⋯+ )∗ ≤ ≤ ⋯ ≤ ℎ : ( + +⋯+ ) ≤ ( + + ⋯+ )( + + ⋯+ )

(Chebyshev’s inequality) 4) If . ≥ 0( = 1; 2; . . ; , = 1; 2; … ; ) then we have: ( . + . + ⋯+ . )( . + . + ⋯+ . ) … ( . + . + ⋯+ . ) ≥ ≥ . . … . + . . … . + ⋯+ . . … . (Hölder’s inequality)

Ex: ∗ ( + )( + ) ≥ √ + √ ∀ , , , ≥ 0



∗ ( + + )( + + )( + + ) ≥ √ + + ∀ , , , , , , , , ≥ 0

→

∗ + + ⋯+ ≥ + + ⋯+ > 0( = 1; 2; … ; ); ∈ , ≥ 2

∗+ +⋯+

≥+ +⋯+

ℎ ( )

∗1

+1

+ ⋯+1≥ ( + + ⋯ ) > 0( = 1; 2; … ; ) →

1+

1+ ⋯+

1≥ ( + + ⋯+ )

→ Let and > 0( = 1; 2; … ; ) and real , > 0 such that + = . Then we have:

+ + ⋯+ ⋅ + +. . + ≥ + + ⋯+ (General Hölder’s inequality) 5) Let , > 0( = 1; 2; … ; ) and any > 1. Then we have:

+ + … ( + ) ≥ ( + + ⋯+ ) + ( + + ⋯+ ) (Minkovsky’s inequality) → similarly for 3 variables , , > 0.

6) For any ≥ −1 we have: ∗ (1 + ) ≥ 1 + ≥ 1 ≤ 0

∗ (1 + ) ≤ 1 + 0 ≤ ≤ 1 (Bernoulli’s

inequality) 7) For any positive integer and , , ≥ 0 we have:

( − )( − ) + ( − )( − ) + ( − )( − ) ≥ 0 (Schur’s inequality) → This is also true for real ≥ 1 and equality happens iff = = or ( ; ; )~(0; ; ) with > 0. Δ Case = 1 – Schur deg 3: All forms: ( , , will be discussed in later part) ∗ + + + 3 ≥ ( + ) + ( + ) + ( + ) → ( + + ) + 9 ≥

≥ 4( + + )( + + ) → + + +9+ + ≥ 2( + + ) →

→ ( + + )( + + + + + ) ≥ 3[ ( + ) + ( + ) + ( + )] ∗ ≥ ( + − )( + − )( + − ) (Well – known result) ∗ ( − ) ( + − ) + ( − ) ( + − ) + ( − ) ( + − ) ≥ 0 ∗ 3( + + ) ≥ ( + + )[2( + + ) − − − ] ∗ 4( + + ) + 15 ≥ ( + + ) * + + +

( )( )( )≥ 2

∗ + + + 3 ≥ + + + + + for , , > 0: = 1. ∗ ( + + )( + + ) ≥ ( + + )( + + ) Δ Case = 2 – Schur deg 4: All forms: ∗ + + + ( + + ) ≥ ( + ) + ( + ) + ( + ) → → + + + 2 ( + + ) ≥ ( + + )( + + )

∗ 2( + + ) − ( + + ) ≤6 ( + + )

+ + + + + ≤9+ +

* [( − )( + − )] + [( − )( + − )] + [( − )( + − )] ≥ 0 → Let , , , , , ≥ 0. Then we have:

( − )( − ) + ( − )( − ) + ( − )( − ) ≥ 0 iff ≥ ≥ and: ∗ ≥ ⋎ ≥ ∗ + ≥ ∗ + ≥ ∗ √ + √ ≥ ∗ ≥ ⋎ ≥ (General Vornicu – Schur inequality)



8) ( )

+( )

+( )

≥( )

for , , ≥ 0, no 2 of which are 0. (Iran 96

inequality) 9) ( + + ) ≥ 3( + + ) (Vasile’s inequality) → The equality happens iff

= = and also for ( ; ; ) = ⋅ sin ; ⋅ sin ; ⋅ sin or any cyclic permutation. 10) Let and ( = 1; 2; … ; ) such that: ∗ ≥ ≥ ⋯ ≥ ≥ 0, ≥ ≥ ⋯ ≥ ≥ 0 ∗ ≥ ; + ≥ + ; … ; + + ⋯+ ≥ + + ⋯+ ∗ + + ⋯+ + = + + ⋯+ + For ≥ 0 we have: ∑ … ≥ ∑ … , where ( ; ; … ; ) are all the permutations of (1; 2; … ; ) (Muirhead’s inequality) E.g: + ≥ ( + ) → + ≥ +

+ + ≥ + + → + + ≥ + + + + ≥ ( + + ) → + + ≥

≥ + + b) For senior and higher classes: (it’s very hard to express the real form of these inequalities so I will try my best.)

1) ∗ If ≥ ≥ ⋯ ≥ ; ≥ ≥ ⋯ ≥≤ ≤ ⋯ ≤ ; ≤ ≤ ⋯ ≤ and ( ; ; … ; ) is an arbitrary

permutation of (1; 2; … ; ) then: + + ⋯+ ≥ + + ⋯+ . ∗ If ≥ ≥ ⋯ ≥ and ≤ ≤ ⋯ ≤ then:

+ + ⋯+ ≤ + + ⋯+ ; ( + + ⋯+ ) ≤ ( + + ⋯+ )( + + ⋯+ )

(Rearrangement inequality) 2) * Convex function: If , ≥ 0 such that + = 1 then ( ) is called a convex function on ( ; ) ⊂ iff ∀ ; ∈ we have: ( + ) ≤ ( ) + ( )

* Concave function: If , ≥ 0 such that + = 1 then ( ) is called a concave function on ( ; ) ⊂ iff ∀ ; ∈ we have: ( + ) ≥ ( ) + ( )

* If ( ) is a convex function on interval ⊂ then for any ∈ ( = 1; 2; … ; ) we have: ⋯ ≤ ( ) ( ) ⋯ ( ) (Classic Jensen’s inequality)

* If ( ) is a convex function on interval ⊂ then for any ∈ ( = 1; 2; … ; ) and > 0 we have: ( ) + ( ) + ⋯+ ( )

+ + ⋯+ ≥+ + ⋯+

+ + ⋯+

And if ( ) is a concave function then the inequality is reversed. (General Jensen’s inequality) →∗ If ( ) is a convex and continuous function on interval ⊂ then for any ∈ ( = 1; 2; … ; ) and ∈ (0,1) such that + + ⋯+ = 1 we have: ( ) + ( ) + ⋯+ ( ) ≥ ( + + ⋯+ ). And if ( ) is a concave

function then the inequality is reversed. * The classic inequality is a special case from the general one with = = ⋯ = * Let and ( = 1; 2; . . ; ) ∈ ( ⊂ ) such that: _ ≥ ≥ ⋯ ≥ ; ≥ ≥ ⋯ ≥



_ ≥ ; + ≥ + ; … ; + + ⋯+ ≥ + + ⋯+ _ + + ⋯+ + = + + ⋯+ + If ( ) is a convex function on then we have:

( ) + ( ) + ⋯+ ( ) ≥ ( ) + ( ) + ⋯+ ( ) (Karamata’s inequality) * If ( ) is a convex function on ⊂ then for ∈ ( = 1; 2; … ; ) we have:

( ) + ( ) + ⋯+ ( ) + ( − 2)+ + ⋯+

≥

≥ ( − 1)[ ( ) + ( ) + ⋯+ ( )] where + = ⋯ ( = 1; 2; … ; ) (Popoviciu’s inequality) 3) Define = + + , = + + , = with , , are any real numbers. If

= − 3 then we have: ≤ ≤ → The minimum and maximum happens iff 2 of 3 variables ; ; are equal. 4) * Let ( ; ; ) be a symmetric polynomial of degree 3 with , , ≥ 0. Then:

( ; ; ) ≥ 0 ⇔ (1; 1; 1); (1; 1; 0); (1; 0; 0) ≥ 0 (SD3 theorem) * Let ( ; ; ) be a cyclic homogeneous polynomial of degree 3 with , , ≥ 0. Then:

( ; ; ) ≥ 0 ⇔ (1; 1; 1) ≥ 0; ( ; ; 0) ≥ 0 (CD3 theorem) → Let ( ; ; ) be a cyclic homogeneous polynomial of degree ( = 3; 4; 5) with

, , ≥ 0. Then ( ; ; ) ≥ 0 ⇔ ( ; 1; 1) ≥ 0 and (0; ; ) ≥ 0. 5) (S.O.S technique) Define = ( − ) + ( − ) + ( − ) , where ; ; are functions with variables ; ; . Then ≥ 0 iff: * ; ; ≥ 0 * ≥ ≥ ; ≥ 0; + ≥ 0; + ≥ 0 * ≥ ≥ ; ≥ 0; ≥ 0; + 2 ≥ 0; + 2 ≥ 0 * ≥ ≥ ; ≥ 0; ≥ 0; + ≥ 0 * + + ≥ 0; + + ≥ 0 → Consider ( ; ; ) = ( − ) + ( − )( − ) ≥ 0 (*) * If ( ; ; ) is symmetric then to prove (*) is true, we assume that ≥ ≥ or

= min{ ; ; } or = max{ ; ; } and prove that , ≥ 0. * If ( ; ; ) is cyclic then to prove (*) is true, we assume that = min{ ; ; } or

= max{ ; ; } and prove that , ≥ 0. (S.S technique) c) Some identities: (*) Some useful identities in inequality that can be proved by S.O.S, S.S technique:

1) + − 2 = ( − ) ; + − 2 = ( )

→ ( + + ⋯+ )− ( + + ⋯+ ) =∑ −

( + + ⋯+ ) + ( + + ⋯+ )

For = 2: 2( + ) − ( + ) = ( )( )

For = 3: 3( + + )− ( + + ) = ( ) ( ) ( )( ) ( )

= 2 ( ) ( )( )( ) ( )

→ ( + + )1

+1

+1− 9 =

( − )+

( − )+

( − )=

2( − )+

1+

1( − )( − )

2) ( + + ) − 3( + + ) = + + − ( + + ) =

=12

[( − ) + ( − ) + ( − ) ] = ( − ) + ( − )( − )

3) + − ( + ) = ( + )( − )



→ + + + + + −32 =

( − )2( + )( + ) +

( − )2( + )( + ) +

( − )2( + )( + ) =

=( − )

( + )( + ) +( + + 2 )( − )( − )

2( + )( + )( + )

→ 3( + + ) − ( + + )( + + ) = ( + )( − ) + ( + )( − ) + ( + )( − ) = 2( + )( − ) + ( + + 2 )( − )( − )

4) ( + + )( + + )− 9 = ( + )( + )( + ) − 8 = = ( − ) + ( − ) + ( − ) = 2 ( − ) + ( + )( − )( − )

5) + + − 3 = ( + + )[( − ) + ( − ) + ( − ) ] = = ( + + )[( − ) + ( − )( − )]

6) + + − 3 = [( − ) (3 + − ) + ( − ) (3 + − ) + ( − ) (3 + − )] =

=( − )

+( − )( − )

7) + + − ( + + ) = ( ) + ( ) + ( ) = + ( − ) + ( − )( − )

8) + + + 3 − ( + ) − ( + )− ( + ) =

=12

[( + − )( − ) + ( + − )( − ) + ( + − )( − ) ]

= ( + − )( − ) + ( − )( − ) 9) + + − ( + + ) =

=1

2{( − ) [( + ) + ] + ( − ) [( + ) + ] + ( − ) [( + ) + ]}

=( + ) +

( − ) +1

+( + )( + )

( − )( − )

10) + + − ( + + ) = + ( − ) + + ( − ) + + ( − ) 11) + + + + + − ( + ) − ( + ) − ( + ) =

=12

[( + )( − ) + ( + )( − ) + ( + )( − ) ]

= ( + )( − ) + ( + + + )( − )( − ) (**) More identities: Actually the first nine identities are rare, so just consider the identity 10 onwards: 1) * ⋅ + ⋅ + ⋅ = + + − 1 with ≠ ≠

* ⋅ + ⋅ + ⋅ = − − − − 1 with ≠ ≠

2) * + + = ( )( )( )( )( )( )

* + + = ( ) ( ) ( )( )( )( )

with ≠ ≠

3) * ⋅ + ⋅ + ⋅ = −1 with ≠ ≠

* ⋅ + ⋅ + ⋅ = ( ) ( ) ( )( )( )( )

4) * ⋅ + ⋅ + ⋅ = + +

* ⋅ + ⋅ + ⋅ = −( + + ) with ≠ ≠ 5) ( − )( + ) + ( − )( + ) + ( − )( + ) = 0

( + )( − ) + ( + )( − ) + ( + )( − ) = −2( − )( − )( − )



6) ( )( )( )

+ ( )( )( )

+ ( )( )( )

= 1 with ≠ ≠

7) ⋅ + ⋅ + ⋅ = −1 with ≠ ≠

8) * ⋅ + ⋅ + ⋅ = + + with ≠ ≠

* ⋅ + ⋅ + ⋅ = −( + + )[ ( − ) + ( − ) + ( − ) ]

9) ⋅ + ⋅ + ⋅ = −1 10) ( + + ) = + + + 3( + )( + )( + ) 11) 2( + + )− ( + + ) = ( + + )( + − )( + − )( + − ) 12) ( + + ) − ( + + ) = ( − )( − )( − )∑ with

, , ∈ and ≥ 2 such that + + = − 2 (?) E.g: for = 2: ( + + ) − ( + + ) = ( − )( − )( − )

= 3: ( + + ) − ( + + ) = ( − )( − )( − )( + + ) = 4: ( + + ) − ( + + ) = ( − )( − )( − )( + + + + + )

13) ( − )( − ) + ( − )( − ) + ( − )( − ) =

=12

[( + − )( − ) + ( + − )( − ) + ( + − )( − ) ]

14) If , , ≠ 0 such that = 1 then: + + = 1

15) + + + = 4 ⇔ 2 + = (4 − )(4− ), etc ⇔

⇔2 +

(2 + )(2 + ) +2 +

(2 + )(2 + ) +2 +

(2 + )(2 + ) =

=(2− )(2 − )

2 + +(2 − )(2 − )

2 + +(2 − )(2 − )

2 + = 2 + + 2 + + 2 + =

= 2 + + 2 + + 2 + = 1 ⇔1

2 + +1

2 + +1

2 + =1

+ + − 2 =

=+ + + 2

2( + + ) − ⇔ ( + + − 2) = (2 − )(2 − )(2 − )

→ From the identity, there exists , , > 0 such that: = 2( )( )

; = 2( )( )

;

= 2( )( )

. And there exists triangle such that: = 2 cos ; = 2 cos ;

= 2 cos 16) + + + 2 = 1 ⇔ + = (1 − )(1− ), etc ⇔

⇔+

(1 + )(1 + ) ++

(1 + )(1 + ) ++

(1 + )(1 + ) =

=(1 − )(1 − )

+ +(1 − )(1 − )

+ +(1 − )(1 − )

+ = 1 ⇔

⇔ + + + + + = 2 ⇔ + + + + + = 1 ⇔

⇔1

+ +1

+ +1

+ =2

+ + − 1 =+ + + 1

+ + − ⇔

⇔ ( + + − 1) = 2(1 − )(1 − )(1 − ) → If we substitute → ; → ; → we will get identity 15, so:

From the identity, there exists , , > 0 such that: =( )( )

; =( )( )

;



=( )( )

. And there exists triangle such that: = cos ; = cos ; = cos .

⨁ Also if we let = ; = ; = with , , > 0 then = , so we have 2 identities 17-18: 17) + + + = 4 ⇔ + + = + + = 1 ⇔

⇔√

2 + + 2 + +√

2 + =+ + √ − 2

18) + + + 2 = 1 ⇔ + + = 2 ⇔ + + = 1 ⇔

⇔√

1 + + 1 + +√

1 + =2

+ + √ − 1

→ From identity 17, there also exists, , , > 0 such that = ; = ; = ,

similarly for identity 18. 19) ( + + ) + ( + + ) + ( + + ) = ( + + ) +

+ ( + + ) + ( + + ) = ( + + )( + + ) ( + + ) + ( + + ) + ( + + ) = ( + + )( + + )

20) ( + + )( + + )( + + ) = + + , where = + + ; = + +

= 3( + + )( + + ) + [( − )( − )( − )]

=34

[ ( + ) + ( + ) + ( + )] +14

[( − )( − )( − )]

=12

[( + + ) ( + + ) + ( + + ) ( + + )]

→ ∗ ( + + ) ( + + ) = ( + 2 )( + 2 )( + 2 ) + [( − )( − )( − )]∗ ( + + ) ( + + ) = (2 + )(2 + )(2 + ) + [( − )( − )( − )]

21) 2( + )( + )( + ) = [ ( + ) + ( + ) + ( + )− 2 ] + +[( − )( − )( − )]

22) 2[ ( − ) + ( − ) + ( − ) ] = [ ( − ) + ( − ) + ( − ) ] + +[( − )( − )( − )]

23) + + = ⇔ = ⇔ = or = or = d) Useful lemmas: In above parts, I’ve showed some of it. In this part I will state more lemmas, maybe a lot but worth it :) • Inequalities with condition about , , (denote = + + ; = + + ;

= ) – part 1: In this part, I will state lemma with familiar conditions, about the „unusual” conditions, I will show in later seasons. d. 1) If , , > 0 such that = then: 1) + + ≥ + + ( , ∈ ; > ) 2) + + ≥ + + ; + + ≥ + + → + + ≥ ( + + − 1) →

→ + + + 3 ≥ + + +1

+1

+1

3) + + ≥ 1 with ∈

4) + + ≤ 1 with ∈



5) * ( )

+( )

+( )

≥

∗( )

+( )

+( )

≥ →( )

+( )

+( )

≥ with , > 0 (*)

6) + + ≤

7) + + ≤ 1 8) Let = + + ; = + + then:

* + 3 ≥ 4 →+ ≥ 4

+ ≥ 4 ⋅ ∗ ≥ 5 − 6

9) + + ≥ 1 → + + ≥ + + ≥ 1 ≥

≥1+ 2 +

1+ 2 +

1+ 2 →

12 + 1 +

12 + 1 +

12 + 1 ≥ 1 ≥

1+ 2 +

1+ 2 +

1+ 2

:1

3 + 1 +1

3 + 1 +1

3 + 1 ≥34 ≥

1+ 3 +

1+ 3 +

1+ 3

10) ( + 1)( + 1)( + 1) ≥ ( + )( + )( + ) ≥ ( + 1)( + 1)( + 1) ≥ 8 11) ( − 1) + 1 + ( − 1) + 1 + ( − 1) + 1 ≤ ( + + ) with

+ + ≥ + + → Special case: + + = + + and without condition = 1.

12) ( )

+( )

+( )

≥

13) ( + + ) ≥ 81( + + ) d. 2) If , , > 0 such that + + = then: 1) + + ≥ 3 ≥ + + 2) + + ≥ 3 for ∈ 3) + + ≥ 3 for , ∈ such that ≥

4) + + ≥ + +

5) + + ≤ + + ≤ + + with all ∈

6) + + ≤ with all ≥ 2

7) + + ≥ with , ∈ such that >

8) ( + + ) ≤ 3 → ( ) ⋅ ( + + ) ≤ 3 with ∈ → + + ≥

≥1

+1

+1≥ + +

9) + + ≥ 1 10) + + + ≤ 4 11) + + ≥ + + 12*) + + ≥ with ≥ 1 d.3) If , , > 0: + + = 3 then: 1) + + ≥ 3 → ( + + ) ≥ 243 2) + + ≥ + +

3) + + ≥ →( )

+( )

+( )

≥



4) + + ≤ + + 5)

( )+

( )+

( )≤ with ≥ 2

→( )

+( )

+( )

≤( )

with , > 0 and

≥ 2( − + ) (?) 6) + + ≤ with ≥ 1

7) + + ≤ 1

8*) + + ≤ 1 d.4) If , , > 0 such that + + = then: 1) + + ≥ + + → √ + √ + √ ≥ + + with ∈ 2) + + ≥ → ( + + ) ≥ 9( + + )

3) + + ≤ 1; + + ≥ 4) + + ≤ 3 5) + + ≤ 2 + → In case , , ≥ 0: the equality happens

if= = = 1

= √2; = 1; = 0

d.5) If , , > 0 such that + + = + + then: 1) + + ≤ 1

2) + + + ≥ 4 → ∗ + + + 1 ≥ 4∗ ( + 1)( + 1)( + 1) ≥ 8

3) + + ≤ 4) ( + )( + )( + ) ≥ 8

5) + + ≥ + + → + + + ≥ 3 + with ≤ 3 6) + + ≥ + 2 d. 6) If , , > 0 such that + + + = then: 1) + + ≥ + + ≥ 3 → + + ≥

→ + + ≥ + +

2) ≥ + + ≥ → ≥ + +

3) + + ≤ 1

4) √ + √ + √ ≤ 3 5) + + ≥ + + 6) + + ≤ + + d.7) If , , > 0 such that + + + = then: 1) + + ≤ + + ≤ 3 → + + ≤ + 2 ≤ + + ≤ 3 2) ≤ + + ≤

( )

3) ( + ) + ( + ) + ( + ) ≤ 6 4) + + ≥ 3 → + + ≥ + +

5) + ≤ + ≤ 2; etc



6) + + + + + ≥ 2( + + ) → + + ≥ √ + √ + √ ≥ + +

d.8) If , , > 0 such that + + = ( + + ) then: ≥ √ → the equality happens if ( ; ; )~( ; ; ) with > 0. d.9) If , , > 0 such that + + = + then: 1) Assume that ( − 1)( − 1) ≥ 0 then: + ≥ 2 → + + + ≥ 4 2) max{ ; ; } ≥ 1; max{ ; ; } ≥ 1 • Inequalities with classic condition (like , , > 0, …) – part 1: d.10) If , , > 0 then: + + ≥

√

d.11) If ; ; … ; > 0 then: + + ⋯+ ≥…

d.12) If , , ≥ 0, no 2 of which are 0 then: + + ≤ 3 ⋅ with ∀ ∈ d.13) If , , > 0 then: + ≥ + + ≥

( )+ 1

d.14) If , , , , , > 0 then: ( + ) + ( + ) + ( + ) ≥ 2 ( + + )( + + )

d.15) If , , , ≥ 0 then: + + + + 2 ≥ + + + + + (Turkevich’s

inequality)→ The equality happens iff = = = and = = = > 0, = 0 or any cyclic permutation. d.16) If , , > 0 then: + + ≥ 4 + + d.17) If , , are sides of a triangle then:

3( + + ) ≥ √ + √ + √ ≥ √ + − + √ + − + √ + − d.18,19) If , , > 0 then:

* + + ≥ ≥

* + + ≥ + + ≥ ( ) d.20) If , , > 0 then:

+ ++

+ ++

+ +≥

+ ++ +

≥+ 2

++ 2

++ 2

d.21) If , , > 0 then: + + ≥ 3

d.22) If , , , , > 0 then: + + ≥

d.23*) If ≥ 0 then: ≥ 1 + + ≥ 1 + → ≥ 1 +

d.24) If > 0∀ = 1; 2; … ; then: ∑⋅

≥ ⋯ ( = )

d.25) If , , > 0 then: + + +( )

≥ 2

d.26) If , , > 0 then: + + + 2 + 1 ≥ 2( + + ) → + + + + 5 ≥ 3( + + )

d.27*) If , , are sides of a triangle then: ( − ) + ( − ) + ( − ) ≥ 0 (IMO 1983)

d.28) If , , > 0 and ∈ then: + + ≥ + + ≥ ⋯ ≥



≥ + + ≥ + + ≥ + +

d.29) If , , > 0 then: ( )( )

+( )( )

+( )( )

≤ 1

d.30) If , , > 0 then: + + ≤√

d.31) If , , are sides of a triangle then: + + ≤ 2( + + ) (the equality happens iff , , are sides of a degenerate triangle.) d.32*) If , , > 0 and ≥ 0 then: + + ≥ + +

d.33) If , , > 0 then: + + +( )( )( )

≥ 4

d.34) If , , ≥ 0, no 2 of which are 0 then: +( )( )( )

≥ 2 (Jack

Garfunkel’s inequality) • Inequalities with classic condition – part 2: In this part I will show some inequalities about variables , , (denote = + + ; = + + ; = ), even Schur’s inequality. Firstly, we have some identities about , , : 1) + + = − 2 2) + + = − 2 3) ( + )( + )( + ) = − → ( + ) + ( + ) + ( + ) = − 3 4) ( + )( + ) + ( + )( + ) + ( + )( + ) = 5) + + = − 3 + 3 6) + + = − 3 + 3 7) + + = − 4 + 2 + 4 8) ( + ) + ( + ) + ( + ) = − 2 − 9) Denote = −4 + + 18 − 4 − 27 , then: * [( − )( − )( − )] =

* + + = √ if ( − )( − )( − ) < 0; √ if ( − )( − )( − ) ≥ 0

* + + = √ if ( − )( − )( − ) < 0; √ if ( − )( − )( − ) ≥ 0

10) ( + ) + ( + ) + ( + ) = −2 + + 4 − 2 − 3 11) ( + ) + ( + ) + ( + ) = − − 4 + 7 + 2 − 3 12) + + = − 6 + 6 + 9 − 12 − 2 + 3 • Some inequalities about the relation of , , : 1) ≥ 3 ; ≥ 3 2) ≥ 9

→ ≥278

( − ) ≥ 3 ≥9

≥ 27

3) + 9 ≥ 4 → ≥ → ≥ max 0; 4) 2 + 9 ≥ 7 5) + 3 ≥ 4 6) + 4 + 6 ≥ 5

7) ≥ → ≥ max 0;



8) ≤ ; ≤ . Combining inequality 3, 6, 8 and we get:

min(5 − )

18 ;− 7 + 13

9 ≥ ≥ max 0;(4 − )

9 ;(4 − )( − )

6

9*) √ ≤ ≤ √ , with = − 3 . This result comes from solving the inequality = −4 + + 18 − 4 − 27 ≥ 0 with variable . → From this result, by AM-GM we have:

* 27 ≥ 9 − 2 − 2 √ = 9 − 2 − ( )( )

≥(9 − 2 )( − 2 )− ( − 3 )[( − 2 ) + ( − 3 )]

( − 2 ) =

=( − 2 )(−3 + 14 − 6 ) − ( − 3 )

( − 2 ) =(4 − )(4 − 10 + 3 )

( − 2 )

* 27 ≤ 9 − 2 + 2 √ = 9 − 2 +( )

≤(9 − 2 )(2 − 3 ) + 2( − 3 ) − + ( − 3 )

(2 − 3 ) =

=(2 − 3 )[2(9 − 2 ) + ( − 3 )(2 − 3 )] + 4 ( − 3 )

2 (2 − 3 )

=(2 − 3 )(−2 + 9 + 9 ) + 4 ( − 6 + 9 )

2 (2 − 3 ) =27 ( − )2 (2 − 3 )

So we have the chain: ≥( )

≥ ≥( )

, more interesting, the third inequality is stronger than Schur deg 3 and 4, since:

( )− =

( )≥ 0 in case 4 ≥

(4 − )(4 − 10 + 3 )27 ( − 2 ) −

(4 − )( − )6 =

(4 − ) ( − 3 )54 ( − 2 ) ≥ 0

Hence we can also conclude that: ≥ max 0;( )

Further more, we can write inequality 9 as: √ ≤ ≤ √

⇔+ √ − 2√

27 ≤ ≤− √ + 2√

27 For my thinking, the , , technique is one of the nicest and hardest techniques to use, due to long calculations, time loss and it requires carefulness in calculation. But the nice thing to say is its benefit in solving hard inequality about symmetric and cyclic expressions. There is a similar technique to , , , that is , , technique (can be found and search in inequality sites and forums)

ELEMENTE DE GEOMETRIA TRIUNGHIULUI IN COORDONATE BARICENTRICE

By Neculai Stanciu-Romania

Articolul vine uşor în completarea programei şcolare din liceu şi are scopul de a pune în evidenţã noi metode de rezolvare a problemelor de geometrie şi de a lãrgi orizontul



matematic al elevilor.În cele ce urmeazã, voi enunţa şase teoreme importante şi voi demonstra numeroase aplicaţii ale acestor teoreme referitoare la unele egalitãţi şi inegalitãţi în triunghi. Teorema 1.Se considerã un triunghi fix ABC şi notãm BC=a,CA=b,AB=c,S=Aria(ABC). Atunci pentru orice ME2 (unde, E2 este planul euclidian)existã şi este unic tripletul ordonat

(x,y,z) R3,x+y+z=1 astfel încât 0 MCzMByMAx şi reciproc,pentru orice triplet ordonat (x,y,z) R3,x+y+z=1 existã şi este unic un punct ME2

astfel încât 0 MCzMByMAx şi în acest caz vom spune cã punctul M are coordonatele baricentrice (x,y,z) în raport cu triunghiul ABC şi vom nota M(x,y,z).Pentru orice XE2 avem

XMXCzXByXAx (demonstraţie în [1]pag.66). Exemple de coordonate baricentrice pentru câteva puncte remarcabile într-un triunghi: 1.1.A(1,0,0),B(0,1,0),C(0,0,1);

);,( 2

,2

,2

.3.1

; 31,

31,

31.2.1

rICinscriscerculuicentrulp

cp

bp

aI

greutatedecentrulG

);,( )(2

,)(2

,)(2

.4.1 aaa rICexinscriscerculuicentrulap

cap

bap

aI

; ,,.5.1 Nagelluipunctulp

cpp

bpp

apN

; )4(

))((,)4(

))((,)4(

))((.6.1 GergoneluipunctulrRr

bpaprRr

apcprRr

cpbp

;,,.7.1 lortocentructgActgBctgCctgActgBctgcH

);,( )2

2sin,2

2sin,2

2sin(.8.1222

ROCscircumscricerculuicentrulS

CRS

BRS

ARO

. ),,(.9.1 222

2

222

2

222

2

Lemoineluipunctulcba

ccba

bcba

aL

Teorema 2.Puterea punctuluiM(x,y,z) E2 faţã de cercul C(O,R),circumscris triunghiului ABC este dat de relaţia:

.);()( 22222 MpROMxyczxbyzaMp cc (demonstraţie în [1] pag.68). Aplicaţii ale teoremei 2.

;9

,9

,9

)(.1.2

2222

22222

222

Rcba

sicbaROGcbaGpc

Demonstraţie:se aplicã teorema2 şi 1.2. ;2,,2,2)(.2.2 22 rRsiRrROIRrIpc

Demonstraţie:rezultã imediat din teorema2 şi 1.3. ;2,2)(.3.2 22

aaaac RrROIRrIp



Demonstraţie:folosim teorema2 şi 1.4. ;2),(4)(.4.2 rRONrRrNpc

Demonstraţie:utilizãm teorema 2 şi 1.5.

.3,),(9

,81coscoscos,

),coscoscos81(,coscoscos8)(.5.2

22222

222

OGOHdeoarececbaROH

iarCBAsi

CBAROHCBARHpc

Demonstraţie:se aplicã teorema 2 şi 1.7.

;4

2)()(.6.2

rRprRrpc

Demonstraţie:se foloseşte teorema 2 şi 1.6.

;34

,,3,3)(.7.2

222

2

22222

2

222

Scba

sicba

abcROLcba

abcLpc

Demonstraţie:rezultã imediat din teorema 2 şi 1.9. Teorema 3.a)Pentru M(x,y,z) E2 existã relaţia:

)(222 MpzMCyMBxMA c b)Pentru orice XE2 existã relaţia:

.

),(222222

2222

xyczxbyzazXCyXBxXAsiMpXMzXCyXBxXA c

(demonstraţie în [1] pag.69) Aplicaţii ale teoremei 3 :

.

,3

;3

;3

3.1.3

2

222222

222222

2

2222222

GpentruXegalitatecu

EXcbaXCXBXA

cbaGCGBGA

EXcbaXGXCXBXAGM

. ,,;

;,2.2.3

2222222

22222

IpentruXegalitatecuEXabccXCbXBaXAabccICbIBaIA

EXabcpXIcXCbXBaXAIM

. ;

;,)(2.3.3222222

22222

a

aaaa

aa

IpentruXegalitatecuaXAabccXCbXBAaIabcCcIBbIIX

EXXIapaXAabccXCbXBIM

).(2, .,

,tan ,:,),,(.4.3

22222

2222222222

RXIpcXCbXBaXAabcIpentruMexempluDeOpentruMdreaptainegalitatesi

MpentruXgasinegalitatecuRXMzXCyXBxXAxyczxbyzaavemEXatunciEzyxDacaM



; , ,,.5.3

222222222

222

OpentruMegalitatecuRzMCyMBxMAsiOMRxyczxbyzazMCyMBxMAMXDaca

.)()()()(3:, ,)(,.6.3222222

222

cyxbxzazyMpMCMBMArezultaundedeyczbMpMAAXDaca

c

c

Teorema 4.Dacã

2,1,),,( 2 kEzyxM kkkk ,atunci distanţa între punctele M1,M2 este datã de relaţia: .))(())(())(( 2

21212

21212

21212

21 cyyxxbxxzzazzyyMM (demonstraţie în [1] pag.70). Aplicaţii ale teoremei 4: Utilizând coordonatele baricentrice (vezi exemplele date)şi teorema 4 obţinem urmãtoarele relaţii:

cdreptunghiABCesteRcbarezultaundedeCBARcbacbaR

CBAROHcbaROG

,8, ),coscoscos1(8)(9

)coscoscos81(.2.4;9

.1.4

2222

22222222

22222

22

;4

)(4.7.4 ;434

31.6.4

;3)(43)(4 .5.4 ;1651659.4.4 ;2;2.3.4

22

22

2

22

222222

222222

RrrprRRSLIrRp

rRprI

rrRRprrRRHIRrrpRrrpGINIOIHNRrROI

.)4()2(2)4(

)2(214.9.4

;)4()52)(2()4()584()4(9

4.8.4

222

222

3232222

2

rRRrRprRR

rRpRH

rRrprRrRrRrrRrRprR

G

Teorema 5. Dacã 2,1,),,( 2 kEzyxM kkkk ,atunci avem:

.)()()(

21 2

12212

12212

12212

21 cyxyxbxzxzazyzyROMOM

(demonstraţie în [1] pag.71.).Aplicaţii ale teoremei 5: Folosind aceastã teoremã obţinem egalitãţi şi inegalitãţi importante printre care cele ce urmeazã:

;,,)(3

,,90)()(3,,3

)(.3.1.5

;,,)3(2

,,90)()3(2,),2(61.2.1.5

90)(

6,,6);(61.1.1.5

)()()(61.1.5

02

2

22

022222

0

222222222

2222

OGONsaucbarRp

iarNOGmrRpsiprRONOG

OGOIsaucbarprRR

iarIOGmrprRRsiRrrpROIOG

AOGm

RcbsiRcbOAOGcbROAOG

cyxbxzazyROMOG



.,,)(

,,90)()(,,21.2.2.5

;,,)5(2,,90)(

)5(2,),(215.1.2.5

);(21)(.2.5

02

22

0

22222

OIOAsaucbarcbaR

iarAOImrcbaRsibcRrROAOI

ONOIsaucbarprRRiarIONm

rprRRsirpRrRONOI

zabycaxbcrRROMOI

Teorema6. Dacã

;)()()(21

)()()(214)

:,2,1,),,(

21221

21221

21221

21212121

2

cyxyxbxzxzazyzy

cayybcxxabzzRrIMIMa

avemkEzyxM kkkk

(demonstraţie în [1],pag.73)

;)()()(21

)()()(61)(

92)

21221

21221

21221

221

221

221

22221

cyxyxbxzxzazyzy

czzbyyaxxcbaGMGMb

(demonstraţie în [1],pag.74)

....)()()(21)() 2

1221212121 azyzyyyzzzyMpMMMMc c

(demonstraţie în [1],pag.75) Observaţie.Cu ajutorul acestor relaţii remarcabile se pot determina ,în particular,produse scalare,distanţe,egalitãţi şi inegalitãţi utilizând puncte din mulţimea: LNIIIIHGOCBA cba ,,,,,,,,,,,, asociatã unui triunghi ABC.(Exerciţiu!). Mai fac observaţia cã particularizãri şi unele extinderi ale coordonatelor baricentrice sunt abordate în lucrãrile [2] , [3] şi articolele din [4] şi [5] .Un fapt care motiveazã studiul coordonatelor baricentrice este legãtura acestora cu calculul vectorial recent (relativ) introdus în programele şcolare IX-XII . Bibliografie [1] .V.Nicula,Geometrie planã,Ed.Gil,2002. [2].N.Teodorescu,ş.a.,Culegere de probleme pentru concursurile de matematicã, vol.5, S.S.M.R, Bucureşti,1977. [3]. M.Craioveanu,I.D.Albu,Geometrie afinã şi euclidianã,Ed.Facla,Timişoara,1982. [4] .T. Bârsan, Recreaţii matematice, nr. 1 / 2002; [5] .C. Coandã, Gazeta matematicã, nr. 8 / 2005; [6].N.Stanciu, Matematicǎ gimnaziu & liceu, Editura ’’Rafet’’, Rm. Sãrat, 2007



FEW APPLICATIONS OF COCEA’S INEQUALITY

By Daniel Sitaru, Claudia Nănuți-Romania

Abstract: In this paper are developed a few applications of a 1993’s inequality of Romanian

mathematician C. Cocea.

Keywords: circumcentre, centroid, orthocentre, incentre, Nagel, Lemoine

Cocea’s inequality

Let be a point in ( ), where , , necollinears. If = ; = ; = then:

+ + ≥ (*)

Proof:

If is the origin of the system of rectangular axis lets denote , , the affixes of ,

respectively . ( ); ( ); ( ); , , different in pairs

1 = |1| = ( − )( − ) + ( − )( − ) + ( − )( − ) ≤

≤ ( − )( − ) + ( − )( − ) + ( − )( − ) =

=| | ⋅ | |

| − | ⋅ | − | +| | ⋅ | |

| − | ⋅ | − | +| | ⋅ | |

| − | ⋅ | − |

| − | ⋅ | | ⋅ | | + | − | ⋅ | | ⋅ | | + | − | ⋅ | | ⋅ | | ≥ | − | ⋅ | − | ⋅ | − |

+ + ≥

Application 1

In Δ the following relationship holds:

≥ 2 (Euler’s inequality)

Solution:

Let = 0 (circumcenter) in (*)

= = = = = =

⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ ≥

( + + ) ≥ 4



⋅ 2 ≥ 4 ⇒ 2 ≥ 4 ⇒ ≥ 2

Application 2


4( + + ) ≥ 9

Solution:

Let = (centroid) in (*)

=23 ; =

23 ; =

23

⋅23 ⋅

23 + ⋅

23 ⋅

23 + ⋅

23 ⋅

23 ≥

4( + + ) ≥ 9

Application 3


sinsin sin

+sin

sin sin+

sinsin sin

≥2

Solution:

Let = (incentre) in (*)

=sin

; =sin

; =sin

⋅sin

⋅sin

+ ⋅sin

⋅sin

+ ⋅sin

⋅sin

≥

sin sin+

sin sin+

sin sin≥

2 sinsin sin

+2 sinsin sin

+2 sin

sin sin≥

4

sinsin sin

+sin

sin sin+

sinsin sin

≥4⋅ 2 =

2

Application 4


sin |cos cos | + sin |cos cos | + sin |cos cos | ≥ 2

Solution:

Let = (orthocenter) in (*)



= 2 |cos |; = 2 |cos |; = 2 |cos |

⋅ 2 |cos | ⋅ 2 |cos | + ⋅ 2 |cos | ⋅ 2 |cos | + ⋅ 2 |cos | ⋅ 2 |cos | ≥

|cos cos | + |cos cos | + |cos cos | ≥ 4

2 sin ⋅ |cos cos | + 2 sin |cos cos | + 2 sin |cos cos | ≥44 =

sin ⋅ |cos cos | + sin |cos cos | + sin |cos cos | ≥ 2

Applicaton 5


(( − ) + 4 )(( − ) + 4 )( , , )

≥

Solution:

Let = (Nagel’s point) in (*)

= ( − ) + 4 ; = ( − ) + 4 ; = ( − ) + 4

⋅ ⋅( , , )

≥

(( − ) + 4 )(( − ) + 4 )( , , )

≥

Application 6


+ + ≥( + + )

4

Solution:

Let = (Lemoine’s point) in (*)

=2+ + ; =

2+ + ; =

2+ +

⋅2+ + ⋅

2+ +

( , , )

≥

4 ( + + )( , , )

≥

+ + ≥( + + )

4

In all applications equality holds if = = (equilateral triangle)



Refferences:

[1] M. Becheanu - “Mathematical Olympiads” – “Gil” Publishing House – Zalău -1997

[2] Mihály Bencze; Daniel Sitaru – “699 Olympic Mathematical Challenges” – “Studis”

Publishing House – Iași – 2017

[3] Mihály Bencze; Daniel Sitaru – “Quantum Mathematical Power” – “Studis” Publishing


[4] Mihály Bencze; Daniel Sitaru – “Olympic Mathematical Energy” – “Studis” Publishing


[5] George Apostolopoulos; Daniel Sitaru – “The Olympic Mathematical Marathon” –

“Cartea Românească” Publishing House – Pitești – 2017

[6] Romanian Mathematical Magazine – Interactive Journal – www.ssmrmh.ro

ABOUT PROBLEM UP.147. ROMANIAN MATHEMATICAL MAGAZINE-AUTUMN EDITION-2018

By Marin Chirciu-Romania

1) In ABC 2 2 22 2 2 9

42 2 2 2 2 2

a b ca b cr r r

B C C A A Btg tg tg tg tg tg

.

Proposed by Hoang Le Nhat Tung, Hanoi, Vietnam Soluție: Demonstrăm rezultatul ajutător: Lemă.

2) In ABC

3 22 2 2 4 12

2 2 2 2 2 2

a b c R r Rpr r rB C C A A B rtg tg tg tg tg tg

.

Demonstrație.

Folosind aSr

p a

și

2 2B C p atg tg

p

obținem:



2

2 3 32 222 2

3 3

4 12 4 121

2 2

a

Sp a R r Rp R r Rpr S p S pB C p a S rp atg tg

p

.

Să trecem la rezolvarea inegalității din enunț. Folosim Lema și 2 2 2 2 22 4a b c p r Rr inegalitatea se scrie:

3 2

2 24 12 9 2 44

R r Rpp r Rr

r

3 2 22 4 9 4 3 8 3R r r R r p R r , care

rezultă din inegalitatea lui Gerretsen: 2 2 24 4 3p R Rr r . Rămâne să arătăm că:

3 2 2 22 4 9 4 3 4 4 3 8 3R r r R r R Rr r R r 3 2 2 38 9 12 4 0R R r Rr r

2 22 8 7 2 0R r R Rr r , evident din inegalitatea lui Euler 2R r .

Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă. Inegalitatea poate fi întărită:

3) In ABC 2 2 2 3 3 327

42 2 2 2 2 2

a b cr r r a b cB C C A A B a b ctg tg tg tg tg tg

.

Marin Chirciu Soluție: Folosind Lema și 3 3 3 2 22 3 6a b c p p r Rr inegalitatea se scrie:

3 2 22 2 3 64 12 274 2

p p r RrR r Rpr p

3 2 24 4 27 6 3 3 16 9R r r R r p R r

care rezultă din inegalitatea lui Gerretsen: 2 2 24 4 3p R Rr r . Rămâne să arătăm că:

3 2 2 24 4 27 6 3 3 4 4 3 16 9R r r R r R Rr r R r

3 2 2 332 54 21 2 0R R r Rr r 2 22 32 10 0R r R Rr r , evident din

inegalitatea lui Euler 2R r . Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă. Inegalitatea 3) este mai tare decât inegalitatea 1):

4) In ABC

2 2 2 3 3 3

2 2 227 94 4

2 2 2 2 2 2

a b cr r r a b c a b cB C C A A B a b ctg tg tg tg tg tg

.

Soluție:

Se folosește inegalitatea 3) și 3 3 3

2 2 227 94 4

a b c a b ca b c

3 3 3 2 2 23 a b c a b c a b c



32 a ab a b , care rezultă din 3 3a b ab a b 2 0a b a b și analoagele. Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă. Să obținem o inegalitate de sens contrar:

5) In ABC

2 2 2

2 281 9 328

2 2 2 2 2 2

a b cr r r R R rB C C A A B rtg tg tg tg tg tg .

Marin Chirciu Soluție: Folosind Lema inegalitatea se scrie:

3 2

2 24 12 81 9 328

R r Rp R R rr r

, care rezultă din inegalitatea lui Euler

2Rr și

inegalitatea lui Mitrinovic 2 227p r . Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă. Se poate scrie dubla inegalitate:

6) In ABC : 2 2 2

2 2 227 815 4 9 322 8

2 2 2 2 2 2

a b cr r r RRr r R rB C C A A B rtg tg tg tg tg tg .

Marin Chirciu Soluție: Se folosesc 3) , 5) și inegalitatea lui Gerretsen 2 216 5p Rr r .Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă: În același registru se pot propune:

7) In ABC

2 2 2

4

2 2 2 2 2 2

a b cr r r r R rB C C A A Bctg ctg ctg ctg ctg ctg .

Soluție:

Folosind aSr

p a

și

2 2B C pctg ctg

p a

obținem:

2

22 2 21 4 4

2 2

a

Sp ar S S R r r R rB C p p p a p Sctg ctgp a

.

8) In ABC : 2 2 2

2 992

2 2 2 2 2 2

a b cr r r Rrr B C C A A Bctg ctg ctg ctg ctg ctg .

Marin Chirciu



Soluție:

Se folosește identitatea 2

4

2 2

ar r R rB Cctg ctg și inegalitatea lui Euler 2R r .

Egalitatea are loc dacă și numai dacă triunghiul este echilateral. 9) In ABC

2 2 22

2 2 23

2 2 2

a b cr r r pA B Ctg tg tg .

Soluție:

Folosind aSr

p a

și

2

2p b p cAtg

p p a

obținem:

2

222 2 3 2

22

1 3 3

2

a

Sp ar S p r p pA p b p c p a p b p c r ptg

p p a

.

10) In ABC 2 2 2 2

2

2 2 2

81814

2 2 2

a b cr r r Rr A B Ctg tg tg .

Marin Chirciu

Soluție: Se folosește identitatea2

2

23

2

ar pAtg și inegalitatea lui Mitrinovic

22 2 2727

4Rr p .

Egalitatea are loc dacă și numai dacă triunghiul este echilateral. 11) In ABC

44 22 2 2

22 2 2

2 16 4 4

2 2 2

a b c p p R R r R rr r rA B C pctg ctg ctg

.

Soluție: Folosind aSr

p a

și

2

2p p aActg

p b p c

obținem:

2

2 44 22 2 2 2

3 2 32

2 16 4 4

2

a

Sp a p b p c p p R R r R rr S r p

A p p a p p r pp actgp b p c

44 2

2

2 16 4 4p p R R r R rp

.



12) In ABC 2 2 2

2 2 2

92

2 2 2

a b cr r r RrA B Cctg ctg ctg .

Marin Chirciu Soluție:

Folosind identitatea 44 22 2 2

22 2 2

2 16 4 4

2 2 2

a b c p p R R r R rr r rA B C pctg ctg ctg

inegalitatea se scrie: 44 2

2

2 16 4 4 92

p p R R r R r Rrp

4 2 2 22 4 128 41 4R r p R Rr p , care

rezultă din inegalitatea Blundon-Gerretsen

22 2 4

16 52 2

R R rRr r p

R r

.

Rămâne să arătăm că:

24 2 24

2 4 128 41 4 16 52 2

R R rR r R Rr Rr r

R r

2 223 44 4 0R Rr r

2 23 2 0R r R r , evident din inegalitatea lui Euler 2R r . Egalitatea are loc dacă și numai dacă triunghiul este echilateral.

ABOUT PROBLEM 3675-CRUX MATHEMATICORUM


1) Let ,a b , and c be the sides of a triangle and let s be its semiperimeter. Let r and R

denote its iradius and circumradius respectively. Prove that

36

cyclic

b s b c s c Ra s a r

.

Proposed by Michel Bataille, Rouen, France Soluție: În articol voi prezenta o soluție diferita de soluția autorului apărută în Crux Mathematicorum Vol. 38 No.8 din 2012 și voi întări dubla inegalitate din enunț.Voi demonstra rezultatul ajutător: Lema:

2) In ABC :

3 2

2

4 22cyclic

b s b c s c R r s r Ra s a Rs

.

Demonstrație:

cyclic

bc s b s c b s b c s cb s b c s ca s a abc s a s b s c



33 2 3 2

2 2

2 4 2 4 24 2

r R r s r R R r s r RRrs r s Rs

.

Să trecem la rezolvarea problemei propuse. Inegalitatea din stânga, ținând seama de Lemă se scrie: 3 2

2

4 26

2R r s r R

Rs

3 24 14R r p R r ,care rezultă din inegalitatea Blundon-

Gerretsen:

22 4

2 2R R r

pR r

. Rămâne să arătăm că:

23 4

4 142 2

R R rR r R r

R r

2 22 3 2 0R Rr r 2 2 0R r R r , evident

din inegalitatea lui Euler: 2R r .Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Inegalitatea din dreapta, ținând seama de Lemă se scrie:

3 2

2

4 2 32

R r s r R RRs r

32 2 26 2 4p R Rr r r R r ,

care rezultă din inegalitatea Gerretsen: 2 216 5p Rr r .Rămâne să arătăm că:

32 2 216 5 6 2 4Rr r R Rr r r R r 3 2 2 316 23 19 2 0R R r Rr r

2 22 16 9 0R r R Rr r , evident din inegalitatea lui Euler: 2R r .

Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă: Inegalitatea din stânga poate fi întărită:

3) In ABC :

4 53cyclic

b s b c s c ra s a R

.

Soluție: Folosind Lema inegalitatea se scrie: 3 2

2

4 2 4 52 3

R r s r R rRs R

3 23 4 46 11R r p R r ,

care rezultă din inegalitatea Blundon-Gerretsen:

22 4

2 2R R r

pR r

.

Rămâne să arătăm că:

23 4

3 4 46 112 2

R R rR r R r

R r

2 22 6 0R Rr r 2 2 3 0R r R r ,

evident din inegalitatea lui Euler: 2R r . Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă: Inegalitatea 3) este mai tare decât inegalitatea stângă din enunț:

4) In ABC :

4 5 63cyclic

b s b c s c ra s a R

.

Soluție: Vezi inegalitatea 3) și 4 5 63

rR

2R r ( inegalitatea lui Euler).Egalitatea are

loc dacă și numai dacă triunghiul este echilateral. Remarcă: Inegalitatea 3) poate fi și ea întărită:



5) In ABC :

2372cyclic

b s b c s c r ra s a R R

.

Soluție: Folosind Lema inegalitatea se scrie: 3 22

2

4 2 372 2

R r s r R r rRs R R

3 2 2 24 2 8 2R R r p R Rr r ,

22 4

2 2R R r

pR r

(inegalitatea Blundon-Gerretsen).Egalitatea are loc dacă și numai dacă

triunghiul este echilateral. Remarcă. Inegalitatea 5) este mai tare decât inegalitatea 3):

6) In ABC :

23 47 5 62 3cyclic

b s b c s c r r ra s a R R R

.

Soluție: Vezi inegalitatea 5) , inegalitatea 4) și23 47 5

2 3r r rR R R

,

adevărată din inegalitatea lui Euler: 2R r .Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă. Să întărim și inegalitatea din dreapta:

7) In ABC :

3 22cyclic

b s b c s c r Ra s a R r

.

Soluție: Folosind Lema inegalitatea se scrie:

3 2

2

4 2 3 22 2

R r s r R r RRs R r

32 2 24 5 4p R Rr r r R r ,

care rezultă din inegalitatea Gerretsen: 2 216 5p Rr r .Rămâne să arătăm că:

32 2 216 5 4 5 4Rr r R Rr r r R r 2 24 7 2 0R Rr r

2 4 0R r R r , evident din inegalitatea lui Euler: 2R r . Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă. Inegalitatea 7) este mai tare decât inegalitatea inegalitatea dreaptă din enunț:

8) In ABC :

3 2 32cyclic

b s b c s c r R Ra s a R r r

.

Soluție: Vezi inegalitatea 7) și 3 2 32

r R RR r r

, adevărată din inegalitatea lui Euler: 2R r .

Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă: Se poate scrie dubla inegalitate care întărește rezultatul din enunț:

9) In ABC :

23 3 272 2cyclic

b s b c s cr r r RR R a s a R r

.



Marin Chirciu Soluție: Vezi inegalitatea 5) și inegalitatea 7).Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă: Se poate scrie șirul de inegalități:

10) In ABC :

24 3 3 2 36 5 73 2 2cyclic

b s b c s cr r r r R RR R R a s a R r r

.

Soluție: Vezi inegalitatea 6) și inegalitatea 8). Egalitatea are loc dacă și numai dacă triunghiul este echilateral.

ABOUT PROBLEM JP.173 ROMANIAN MATHEMATICAL MAGAZINE-SPRING EDITION 2019


1) Let ABC be a triangle. Prove that 1 1 1 6

sin sin sinR

A B C r .

Proposed by Nguyen Viet Hung-Hanoi-Vietnam Soluție: Demonstrăm rezultatul ajutător: Lemă.

2) In ABC : 2 21 1 1 4

sin sin sin 2p r Rr

A B C rp

.

Demonstrație:

Folosind sin2aAR

, 2 2 4bc p r Rr și 4abc Rrp , obținem:

2 2 2 21 2 1 4 42 2 2sin 4 2

bcR p r Rr p r RrR R RA a a abc Rrp rp

.

Să trecem la rezolvarea problemei din enunț: Folosind Lema inegalitatea de demonstrat se scrie:

2 2 4 62

p r Rr Rrp r

22 2 4 62

p r Rr Rrp r

24 2 2 2

2 2

2 4 4 64

p p r Rr r R r Rr p r



22 2 2 22 16 4 0p p r Rr r R r .

Distingem cazurile: Cazul 1). Dacă 2 22 16 0p r Rr , inegalitatea este evidentă.

Cazul 2). Dacă 2 22 16 0p r Rr , inegalitatea se rescrie:

22 2 2 24 16 2r R r p Rr r p , care rezultă din inegalitatea Blundon-Gerretsen:

22 2 4

16 52 2

R R rRr r p

R r

. Rămâne să arătăm că :

222 2 24

4 16 2 16 52 2

R R rr R r Rr r Rr r

R r

22 3

2 2RrrR r

2 2 3R r R

2R r , (inegalitatea lui Euler). Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă:Să punem în evidență o inegalitate de sens contrar:

3) In ABC : 1 1 1 3

sin sin sinR

A B C r .

Marin Chirciu Soluție: Folosind Lema inegalitatea se scrie:

2 2 4 32

p r Rr Rrp r

22 224 3

2p r Rr R

p

24 2 2 2

22

2 4 43

4p p r Rr r R r

Rp

22 2 2 2 24 12 8 2r R r p R Rr r p ,

care rezultă din inegalitatea Gerretsen: 2

2 2 2 2416 5 4 4 3

r R rRr r p R Rr r

R r

. Rămâne să arătăm că :

2

22 2 2 2 244 12 8 2 4 4 3

r R rr R r R Rr r R Rr r

R r

2 28 12 5r R r R Rr r 2 28 13 6 0R Rr r 2 8 3 0R r R r , evidentă din inegalitatea lui Euler 2R r .Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă: Se poate scrie dubla inegalitate:

4) In ABC : 6 1 1 1 3

sin sin sinR Rr A B C r

.

Soluție: Vezi inegalitățile 2) și 3). Egalitatea are loc dacă și numai dacă triunghiul este echilateral.



ABOUT INEQUALITY IN TRIANGLE-884 ROMANIAN MATHEMATICAL MAGAZINE


1) In ABC 3

2a b cm m m

bc ca ab R .

Daniel Sitaru Soluție.

Folosind 2 abc Rh și a am h obținem 1 1 332 2 2 2

a a a

a a

m m mbc Rh R h R R

.

Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă: Să punem în evidență o inegalitate de sens contrar.

2) In ABC

34

a b cm m mbc ca ab r

.

Marin Chirciu Soluție. Folosind 2 abc Rh obținem

12 2

a a a

a a

m m mbc Rh R h

, (1).

Tripletele , ,a b cm m m și 1 1 1, , ,a b ch h h

sunt invers ordonate. Cu inegalitatea lui Cebâșev

obținem: 1 1 1 9 1 33 3 2 2

aa

a a

m R Rmh h r r

, ( 2).

Din (1) și (2) rezultă concluzia. Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă. Se poate scrie dubla inegalitate:

3) In ABC 3 3

2 4a b cm m m

R bc ca ab r .

Daniel Sitaru,Marin Chirciu



Soluție:Vezi inegalitățile 1) și 2).Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă:Dubla inegalitate poate fi întărită:

4) In ABC

1 7 1 22 2 3 2

a b cr m m m rR R bc ca ab r R

.

Soluție: Inegalitatea din stânga : 1 72 2

a b cm m m rbc ca ab R R

rezultă din:

12 2

a a a

a a

m m mbc Rh R h

, (1).

Cu inegalitatea lui Tereshin 2 2

4ab cm

R

, 2abchR

, 2 2 2 2 2

2b c p r Rr

bc Rr

și

inegalitatea lui Gerretsen: 2 216 5p Rr r obținem: 2 2

2 2 2 2 2 2 2 21 1 2 2 242 2 2 4 4

2

a

a

b cm b c p r Rr p r Rr p r RrR

bch bc Rr Rr RrR

2 2 216 5 2 14 4 7 2 74 4 2 2

Rr r r Rr Rr r R r rRr Rr R R

, (2).

Din (1) și (2) obținem 1 72 2

a b cm m m rbc ca ab R R

.

Inegalitatea din dreapta: 1 23 2

a b cm m m rbc ca ab r R

. Folosind 2 abc Rh obținem

12 2

a a a

a a

m m mbc Rh R h

, (1). Tripletele , ,a b cm m m și 1 1 1, , ,a b ch h h

sunt invers

ordonate. Cu inegalitatea lui Cebâșev obținem:

1 1 1 1 443 3 3

aa

a a

m R rm R rh h r r

, ( 2).

Din (1) și (2) rezultă 1 4 4 1 22 3 6 3 2

a b cm m m R r R r rbc ca ab R r Rr r R

.

Egalitatea are loc dacă și numai dacă triunghiul este echilateral. Remarcă: Dubla inegalitate 4) este mai tare decât 3).

5) In ABC

3 1 7 1 322 2 2 3 2 4

a b cr m m m rR R R bc ca ab r R r

.

Soluție: Vezi 4) și inegalitatea lui Euler 2R r .Egalitatea are loc dacă și numai dacă triunghiul este echilateral.



POLYNOMES AND MATRICES

By Marian Ursărescu-Romania

Within this article we’ll try to solve problems which ask to demonstrate that the determinant of a matrice is positive.First let’s remind the following properties of the numbers: 1. + + ⋯+ = + + ⋯+ ,∀ , , … , ∈ ℂ∗ 2. ⋅ ⋅ … ⋅ = ⋅ ⋅ … ⋅ ,∀ , , … , ∈ ℂ∗ 3. | | = ⋅ ,∀ ∈ ℂ∗ Theorem 1 : 1. det = det ,∀ ∈ ℳ(ℂ) 2. det ⋅ ≥ 0,∀ ∈ ℳ(ℂ) Demonstration: 1. Consider = =

det = ( ) ( ) ( ) … ( )∈

det = ( ) ( ) ( ) … ( )∈

=

= ( )∈

( ) ( ) … ( ) = ( )∈

( ) ( ) … ( ) = det

2. det ⋅ = det ⋅ det = det ⋅ det = |det | ≥ 0 Theorem 2: Consider ∈ ℝ[ ], ( ) = + + , write > 0 and Δ = − 4 < 0. Then det( + + ) ≥ 0,∀ ∈ ℳ (ℝ). Demonstration: If Δ < 0 then has conjugated complex roots: , ∈ ℂ, with = det[( − )( − )] = det[( − )( − )] = det[( − )( − )] ≥ 0 (according to theorem 1) On the other hand: det[( − )( − )] = det[ − ( + ) + ] =

= det + + =1

det( + + )

Consequently det( + + ) ≥ 0.The generalization of theorem 2 Consider ∈ ℝ[ ], ( ) = + + ⋯+ + , > 0 and all its roots are complexe. Then + ⋯+ + ≥ 0,∀ ∈ ℳ (ℝ) Demonstration: If has all roots complex (numbers) and ∈ ℝ[ ] then

( ) = ( − )( − ) … − − and we can use the second theorem for the polynome ( − )( − ), = 1,



Theorem 3: Consider ∈ ℝ[ ], ( ) = + + ,Δ = − 4 < 0,∀ , ∈ ℳ (ℝ) the following statement is true: det[( + )( + )] ≥ 0, , being the roots of the polynome . Demonstration: If Δ < 0 then , ∈ ℂ and =

det[( + )( + )] = det[( + )( + )] = det[( + )( + )] ≥ 0 (according to theorem 1) Notice that if = then the conclusion of the third theorem can be written like that:

det( − + ) ≥ 0 Demonstration: det[( + )( + )] = det[ + ( + ) + ] = = det( − + ) ≥ 0, (we used = and Viète’s relations) Theorem 4: Consider ∈ ℝ[ ], ( ) = + + ,Δ = − 4 < 0,∀ , ∈ ℳ (ℝ) the following statement is true: det[( + + )( + + )] ≥ 0, , being the roots of . Demonstration: If Δ < 0 then , ∈ ℂ and =

det[( + + )( + + )] = det[( + + )( + + )] = = det[( + + )( + + )] ≥ 0 (according to the first theorem) If in addition, = then the conclusion of the fourth theorem can be written like that:

det[ + ( + ) − ( + ) + ( − 2 ) ] ≥ 0 Demonstration: det[( + + )( + + )] =

= det[ + ( + ) + ( + )( + ) + ( + ) ] = = det[ + ( + ) − ( + ) + ( + 2 ) ](we used = and Viète’s relations) Theorem 5: Consider ∈ ℝ[ ], ( ) = + + ,Δ = − 4 < 0,∀ , ∈ ℳ (ℝ) the following statement is true: det[( + + )( + + )] ≥ 0, , being the roots of . Demonstration: If Δ < 0 then , ∈ ℂ and =

det[( + + )( + + )] = det[( + + )( + + )] = = det[( + + )( + + )] ≥ 0 (according to the first theorem)

If = , = , = , then the conclusion of this theorem can be written this way: det[ + ( + ) − ( + ) + ( + 2 ) ] ≥ 0 Demonstration: det[( + + )( + + )] =

= det[ + ( + ) + ( + )( + ) + ( + ) ] = = det[ + ( + ) − ( + ) + ( + 2 ) ] ≥ 0

(we used = , = , = and Viète’s relations). Now let’s solve some problems related to these theorems.

1. Show that det( + ) ≥ 0,∀ ∈ ℳ (ℝ) Demonstration: We use theorem 2 for ( ) = + 1, whose roots are = , = − 2. Show that det( + + ) ≥ 0,∀ ∈ ℳ (ℝ) Demonstration: We use theorem 2 for ( ) = + + 1, with the roots

=−1 + √3

2 , =−1 − √3

2

3. If ∀ , ∈ ℳ (ℝ) so that = , then det( + ) ≥ 0



Demonstration: We use the third theorem for ( ) = + 1, where + = 0, = 1 4. If , ∈ ℳ (ℝ) so that = , then det( + 2 + 2 ) ≥ 0 Demonstration: Using the third theorem we consider ( ) = + 2 + 2, + = −2, = 2 5. If , ∈ ℳ (ℝ), + = , then det( + ) ≥ 0 Demonstration: Let’s show first that = , + = , we multiply right by ⇒ + = (1) + = , we multiply left by ⇒ + = (2)

+ = ℎ = −+ = = −

⇒ =

det( + ) = det[( + )( − + )] = det( + ) det( − + ) = = det( − + ) ≥ 0 (using theorem 3 for ( ) = − + 1)

6. If , , ∈ ℳ (ℝ), = , = , = then

det( + + − − − ) ≥ 0 Demonstration: We use theorem 5 for ( ) = + + 1, whose roots are , ∈ ℂ with

+ = −1, = 1 and + = −1. 7. If , ∈ ℳ (ℝ), = then det(2 + 2 − 3 − − + ) ≥ 0 Demonstration: We use theorem 4 for ( ) = − + 2, with the roots , ∈ ℂ, +

= 1, = 2 and = 2 and + = −3. 8. If , , ∈ ℳ (ℝ), = , = , = , then

det( + ( + )( + ) + 2 ( + ) + 2( + ) ) ≥ 0,∀ , ∈ ℝ Demonstration: Using theorem 5 we consider ( ) = + 2 + + ,Δ = −4 ≤ 0, and the roots , ∈ ℂ , + = 2 , = + , + = 2( + ). 9. If , ∈ ℳ (ℝ), = , then:

det( + 2( + )( + ) + 2( + )( + ) + 8 ) ≥ 0,∀ , ∈ ℝ Demonstration: We use theorem 4 for, ( ) = − 2( + ) + 2 + 2 , Δ = −4( − ) ≤ 0, , ∈ ℂ, with + = 2( + ) , + = 2( + ) 10. If ∈ ℳ (ℝ), then det( + + + ⋯+ ) ≥ 0,∀ ∈ ℕ∗ Demonstration: We use the generalization of theorem 2 for ( ) = 1 + + + ⋯+ which has only complex roots BIBLIOGRAPHY: Maths Review

PROPOSED PROBLEMS

5-CLASS-STANDARD



V.1. Prove that 1x abc if and only if 1x x ab c .Proposed by Marin Chirciu – Romania

V.2. Prove that 4 3 7 57 3n n , where n N . Proposed by Marin Chirciu – Romania

V.3. Let be the number 2 1 1 2 3 ... 2 , A n n n N . Prove that

22 1 , A n n n N . Prove that if 25 10, n k k N , then A can be written as a sum of

five perfect squares, nonzero and distinct, for k ≥ 3.

Proposed by Marin Chirciu – Romania

V.4. Find all natural numbers having maximum six digits that have the digits directly proportional with their rank from the right to the left.

Proposed by Daniel Sitaru-Romania

V.5. Prove that it exists an infinity of natural numbers , ,x y z for which the number

13 13 13x y z to be a perfect cube. Proposed by Marin Chirciu – Romania V.6. Find the natural numbers n such that the sum n ...21 to be a number formed from two identical digits. Proposed by Titu Zvonaru, Neculai Stanciu-Romania

V.7. Let be the natural numbers cba ,, , zyx ,, which verifies the relationships yxxya , zxxzb , zyyzc . Prove that the product of the successors of the numbers

cba ,, is a perfect square. Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

V.8. Find all the numbers ba879 divisible with 72. Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.9. Find all the numbers abcde for which edcbaabcde 4 .

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.10. Prove that the nonzero natural numbers cba ,, satisfy the relationship

17202512943 2 cba , then they are prime.

Proposed by Titu Zvonaru, Neculai Stanciu-Romania

V.11. Prove that the number 194800…007171, which has in total 2016 zeros, is composed. Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

V.12. If to different letters correspond different digits, then find all the numbers abc knowing that the arithmetic means of the numbers abc , acb , bac , bca , cab and cba is abc .




V.13. Solve the equation abcabcxxx )2)(1( .


V.14. If to different letters correspond different digits then find all the numbers abc which verify the relationship accacbbcbaababc .


V.15. If to different letters correspond different digits such that abbbadbaccabdbcdaadcb ,then find abcd .


V.16. If to different letters corresponds different digits and abcd 2 , then find all the perfect squares having the form abcdab .

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania V.17. In how many ways can we weight 10 grams with a balance if we have 8 weights of one gram, 3 of two grams and 4 of three grams?

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania V.18. What age is Bogdan, knowing that his age is one year bigger than eight times the sum of the digits that it is formed?

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania V.19. Prove that if p is odd number and not divisible with 3 then the number 20152 p divides with 24.

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania V.20. How many digit has the number 962 ?

Proposed by Titu Zvonaru, Neculai Stanciu-Romania V.21. Suma a 12 numere naturale consecutive este un număr cu cifre diferite, printre care se află cifrele 1,2,3 și 4. Care este cel mai mic număr posibil dintre cele 12 numere?

Proposed by Ionică Constantin-Romania

V.22. Determinați numărul , știind că 4 + 1026 = Proposed by Ionică Constantin-Romania

V.23. Scrieți numărul 164 ca sumă de patru pătrate perfecte.


V.24. Se consideră numărul = 7 + 7 + 7 + ⋯+ 7 . Arătați că 6 + 7 este pătrat perfect. Proposed by Ionică Constantin-Romania V.25. Câte numere naturale de patru cifre se pot scrie ca sumă a cinci numere naturale consecutive? Proposed by Ionică Constantin-Romania



V.26. Să se arate că dacă 23 ⋅ = + 369, atunci numărul se divide cu 41.

Proposed by Ionică Constantin-Romania All solutions for proposed problems can be finded on the

http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.

6-CLASS-STANDARD

VI.1. Find all the pairs ),( yx of integers which verify the relationship 20182 yxyx . Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VI.2. Let be fedcba ,,,,, real numbers, strictly positive which verify the relationships (i) 16222 cba ; (ii) 36222 fed ; (iii) 24 cfbead .

Find 32

fedcba

. Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VI.3. If 5p is a prime number, then find the reminder of the number division 2p to 12.

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VI.4. If a and b are digits, then solve the equation 2))(( xbaabbaab .


VI.5. Find all the pairs QZyx ),( , which verify the relationship 57

2

22

yx

yxyx.

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania VI.6. Find all the integers a for which aa 82 is a perfect square.


VI.7. Find the set

ZxxZxA 4

8

34918

.

Proposed by Neculai Stanciu,Iuliana Traşcă-Romania



VI.8. Let be the isosceles triangle ABC with ACAB , the point D on the line BC , with C between B and D and the point E the symmetric of the point B towards the line AD . Prove that ECDBAD . Proposed by Titu Zvonaru,Neculai Stanciu-Romania

VI.9. Prove that it doesn’t exists a prime number p such that both numbers 20153 p and

20173 p to be prime. Proposed by Neculai Stanciu,Titu Zvonaru-Romania

VI.10. Fie A= 472018 , arătați că :

a) A nu se poate scrie ca sumă a două cuburi de numere întregi b) A nu se poate scrie ca suma a trei cuburi de numere întregi.

Proposed by Cremeneanu Luiza Lorena, Prunaru Constantina-Romania

VI.11. If ,a bZ such that 25 12a b is a multiple of 11, then 2 3a b is a multiple of 11.

Proposed by Marin Chirciu-Romania

VI.12. Solve in nonzero natural numbers set the following equation 2111111

cbaProposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VI.13. If to different letters correspond different digits, then find all the digits cba ,, which

verify the relationship cbacbaab . Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VI.14. From 100 pupils of a school, that have participated at baccalaureate exam, 70 passed test A, 80 test B, 75 test C and respectively 85 test D. Prove that at least 10 pupils have passed this exam.

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania VI.15. Prove that the number 102400…0081, which has in total 2015 zeros, is composed

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania VI.16. Prove that for any natural number n 20121 n , it exists a natural number msuch that all the numbers 12 nm to be composed.

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania VI.17. Find all the triplets zyx ,, of natural numbers with zyx 0 , which verify the

relationship mzyx 222 *Nm .


VI.18. If 2011

1a , then compute the whole part of the number

201122 ...11...

11

11

aaaaaaS

.




VI.19. If *,, Rcba such that 3 cabcabcba , then find:

201120112011 cba . Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VI.20. Let cba ,, be the nonzero real numbers such that 9 cba and 27 cabcab . Find 201420142014 cba .


VI.21. Arătați că fracția ⋅ ⋅ ⋅…⋅⋅ ⋅ ⋅…⋅

este ireductibilă. Proposed by Ionică Constantin-Romania

VI.22. Fie numerele natural și . Arătați că dacă fracția se simplifică prin 17, atunci și

fracția se simplifică cu 17. Proposed by Ionică Constantin-Romania VI.23. Arătați că nu există numere naturale , , pentru care + + = 18 și + + = 107. Proposed by Ionică Constantin-Romania

VI.24. Se consideră numerele raționale = și = . Arătați că + > 2,7(3).


VI.25. Fie , , ∈ ℕ care nu sunt divizibile cu 3. Arătați că suma + + este divizibilă cu 3. Proposed by Ionică Constantin-Romania VI.26. Numerele + și − sunt direct proporționale cu 12 și 3. Determinați numerele și știind că ⋅ = 135. Proposed by Ionică Constantin-Romania

All solutions for proposed problems can be finded on the


7-CLASS-STANDARD

VII.1. Let , , be positive real numbers satisfying + + = 3. Prove that: ++ +

++ +

++ ≥ 3

Proposed by Pham Quoc Sang-Vietnam



VII.2. Find all pairs (x, y) of integers satisfying the equation

x − (y + 2)x + (y − 1)x + (y + 2)x + y = 2. Proposed by Nguyen Viet Hung-Hanoi-Vietnam

VII.3. Find the numbers , , ∈ ℕ∗ knowing that:

∈ ℕ, ∈ ℕ and ∈ ℕ

Proposed by Gheorghe Alexe, George – Florin Șerban– Romania

VII.4. Let be a trapeze where ∥ ; = ; = ; = ; = ; > .

Prove that Area [ ] < ( )( )( )

. Proposed by Daniel Sitaru – Romania

VII.5. Given , , > 0 and + + = 6, prove + + + + + ≥ 6

Proposed by Nguyen Phuc Tang– Vietnam

VII.6. Prove that the following inequalities hold for all positive real numbers:

a. + + ≥ ⋅

b. ( )

+( )

+( )

≥ ⋅

Proposed by Nguyen Viet Hung–Vietnam

VII.7. Prove that the following inequalities holds for all positive real numbers a, b, c:

a. + + ≥ ( )

b. + + ≥ ( )

Proposed by Nguyen Viet Hung-Vietnam

VII.8. Prove that for all positive real numbers a, b, c :

a(b + c )2a + bc +

b(c + a )2b + ca +

c(a + b )2c + ab ≥

6abcab + bc + ca

Proposed by Hung Nguyen Viet-Vietnam VII.9. Let x, y, z > 0 be positive real numbers. Then

1x + y +

1y + z +

1z + x ≥

4 3xyz(x + y + z)(x + y)(y + z)(z + x)

Proposed by D.M. Bătinețu-Giurgiu-Romania, Martin Lukarevski-Macedonia



VII.10. Prove that for all positive real numbers a, b, c:

a. + + ≥ + ( )( )

b. + + ≥ + ( )( )

Proposed by Nguyen Viet Hung-Vietnam

VII.11 Prove that if ∈ (0,∞); ∈ 1, ; ∈ ℕ; ≥ 3; = ; ⋅… ⋅ = 1, then

+ + 1

+ +≥ √3

Proposed by Daniel Sitaru– Romania

VII.12. Solve for natural numbers: + + + ⋯+ = 3025

Proposed by Daniel Sitaru– Romania VII.13. We consider the right triangle ABC with the right angle in A . Let D be the contact point of altitude from A on BC and E the bisector’s intersection of the angle ADC with

AC . If )(),( DCNAEM and ADMNF such that 1 kNCME

DCAE , find in

function of k the rapport AFBD .

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania VII.14. Prove that the number = 111 … 1

is composed.


VII.15. Prove that the number 999 … 9

+ 1 999 … 9 000 … 0 is composed.


VII.16. We consider an ABC triangle with the lengths sides 30,34,16 CABCAB and the points M respectively N on the side BC such that 12,4 MNBM . Compute the measure of the angle MAN .


VII.17. We denote “Super – Heron” triangle (do not confuse with “Super – Hero”) a triangle that has the lengths sides consecutive natural numbers and the area also a natural number.



Recent1(2007) was proved the existence of an infinite number of such triangles. Confirm or infirm the existence of “Super – Heron” inscriptible quadrilaterals. (Indication. Use the area formula of the incriptible quadrilateral given by the Indian mathematician Brahamgupta in VII century - A.H: ))()()(( dscsbsasA , where

dcba ,,, are the quadrilateral lengths sides, and s is its semiperimeter) Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania

VII.18. We consider a right-angled triangle with the perimeter P and the area A natural numbers.Prove that: the hypotenuse is a natural number if and only if P is natural even

number and 2P divides A AP 2 .

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania VII.19. Which conditions must meet the triangle’s angle so this can be divided in two isosceles triangles? Find the angles of an isosceles triangle which can be divided in two isosceles triangles. Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania VII.20. Find the biggest triangle with the dimensions natural numbers whose sum is even number for which the area is also a natural number and it is equal with its perimeter.


VII.21. Fie , , numere reale astfel încât au loc simultan relațiile: 5 − 7 + 9 ≥ 0, 7 − 9 + 11 ≥ 0, 9 − 5 − 20 ≥ 0. Arătați că 5 − 28 + 27 = 24.


VII.22. Să se arate că într-un triunghi cu lungimile laturilor , , și care are perimetrul egal cu 15 este adevărată relația √ + − + √ + − + √ + − ≤ 9.


VII.23. Dacă , , , sunt lungimile laturilor unui patrulater și verifică relația: = = = , stabiliți natura patrulaterului.


VII.24. Se știe că lungimile diagonalelor unui romb verifică relația: + = 2√2. Determinați măsurile unghiurilor rombului. Proposed by Ionică Constantin-Romania VII.25. Stabiliți dacă numărul:

= 1 + 2018 ⋅ 1 + 2019 ⋅ 1 + 2020 ⋅ 1 + 2021 ⋅ √1 + 2022 ⋅ 2024 este rațional

sau irațional. Proposed by Ionică Constantin-Romania VII.26. Arătați că numărul = 2019 − 10 ⋅ 2018 − 20 ⋅ 2018 + 15 este pătrat perfect.


1 http://www.math.twsu.edu/~richardson/heronian/heronian.html



VII.27. Să se rezolve ecuația 2 22 2

4a b a a a b b b

abb a

, unde a, b și c sunt strict

pozitive și 2a b . Proposed by Carmen Chirfot-Romania

VII.28. Să se demonstreze că 2 1 1 2 24 a ba b aba b b a

, unde a, b și c sunt numere reale

strict pozitive. Proposed by Carmen Chirfot-Romania

All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical

Magazine-Interactive Journal.

8-CLASS-STANDARD

VIII.1. Solve for integers:

+−2 +

−6 = 2018201720162015

Proposed by Rovsen Pirguliyev-Sumgait-Azerbaijan

VIII.2. If = 2018 , ∈ ℕ then: a.Prove that can not be written as a sum of two perfect cubes. b.Prove that can not be written as a sum of three perfect cubes.

Proposed by Lucian Tutescu,Daniela Iancu– Romania

VIII.3. If , , > 0,√1 + + √1 + + √1 + = 3√2 then:

√1 ++√1 +

+√1 +

≤3√2

2


VIII.4. If , , > 0, + + = 3 then: ∑ ( + 2 ) + ( + 2 ) ≤ 6√3




VIII.5. Solve in 2R the system of equations:

2012)1)(1( 22 yyxx ; yx2012

2011.

Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania VIII.6. If , , > 0, + + = 1 then: ∑( + ) ≥ 4 3


VIII.7. Prove that if a, b, c > 0 then:

ab + c + 2

bc + a + 4

ca + b ≤ 7

ab + c +

2bc + a +

4ca + b

Proposed by Daniel Sitaru– Romania VIII.8. If , , >, + + = 3 then:

+ − 1+ − 1 ≥ + +

Proposed by Daniel Sitaru – Romania

VIII.9. If , , > 0, + + = 3 then:

+ + 1 + + + 1 + + + 1 < 1 + + +


VIII.10. If , , > 0, + + = 1 then:

+ + + 2 2 + + ≥ 3

Proposed by Daniel Sitaru-Romania VIII.11. If , , > 0 then:

++ +

++ +

++ ≤ + +


VIII.12. If , , > 0, = 1 then:

( − 1 + ) ( + ) ≥ 6


VIII.13. If , , > 0then:

1 + ++

1 + ++

1 + +< 1 + + +

Proposed by Daniel Sitaru – Romania VIII.14. Solve for real numbers:



[| |] = |[ ]|[| |] = |[ ]| , [∗] - great integer function

Proposed by Daniel Sitaru – Romania VIII.15. If , , > 0 and + + = 3 prove that:

1+

1≥

18+ +

where ≥ 0. Proposed by Marin Chirciu– Romania

VIII.16. Let , , be positive real numbers. Prove that:

( − + )( + ) +

( − + )( + ) +

( − + )( + ) ≥

316

Proposed by George Apostolopoulos-Greece

VIII.17. Prove that if , , ∈ ℝ then:

(2 − − − + ) ≤ ( + 2)( + 2)( + 2) Proposed by Daniel Sitaru – Romania

VIII.18. If , , > 0; ≥ 1 then:

3 ( + + )( + + ) +

+ ++ + ≥ + 1


VIII.19. If , , > 0 then:

136 + 7 +

136 + 7 +

136 + 7 ≤ 3

Proposed by Marin Chirciu– Romania

VIII.20. Let a, b, c be non-negative such that a + b + c = 3. Prove that:

|(a − b)(b − c)(c − a)| ≤3√3

2

Equality occurs when? Proposed by Nguyen Ngoc Tu-Vietnam

VIII.21. Rezolvați în ℝ ecuația: | − 2| ⋅ | + 5| = | + 2| ⋅ | − 5| Proposed by Ionică Constantin-Romania

VIII.22. Fie , ∈ ℕ∗ astfel încât 61|(5 + 6 ). Arătați că 61|( + ). Proposed by Ionică Constantin-Romania

VIII.23. Fie , , > 0 numere reale astfel încât = 3. Arătați că: 3

(3 + )( + 3 ) +4

(4 + )( + 4 ) +5

(5 + )( + 5 ) ≤14




VIII.24. Rezolvați ecuația: − 1

2019 +− 3

2017 +− 5

2015 +− 7

2013 =− 2019

1 +− 2017

3 +− 2015

5 +− 2013

7

Proposed by Ionică Constantin-Romania VIII.25. Rezolvați în muțimea numerelor întregi ecuația: + − = 2019.

Proposed by Ionică Constantin-Romania VIII.26. Rezolvați în mulțimea numerelor întregi ecuația:

7 + 7 + 8 + 8 + 2018 = 0 Proposed by Ionică Constantin-Romania

VIII.27. Să se demonstreze că 3 3 3 3 3 3

6a b b c c aab bc ca

, unde a, b și c sunt strict pozitive

și 3a b c . Când are loc egalitatea? Proposed by Carmen Chirfot-Romania

VIII.28. Să se rezolve ecuația 2 2 23 3 3 6a b c , unde a, b și c sunt pozitive și

3a b c . Proposed by Carmen Chirfot-Romania

VIII.29. Să se rezolve ecuația 12111222 cba

ccbbaa , unde a, b

și c sunt strict pozitive și 3a b c . Proposed by Carmen Chirfot-Romania


Magazine-Interactive Journal. 9-CLASS-STANDARD

IX.1. If , ∈ ℕ; , , > 0 then: ( + 1) +

1( + ) + ( + 1) +

1( + ) + +

1( + ) +

1( + ) ≥

≥9( + 1)( + 1)4( + + )

Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru-Romania IX.2. If , ∈ ℕ; , > 0 then in Δ the following relationship holds:

( + )( + ( ) ) + ( + ( ) )( + ( ) ) + +( + ( ) )( + ) ≥ ( + 1)( + 1) √3

Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru-Romania



IX.3. Let ∈ 0; and ∈ ℕ∗. Prove: sin

cos +cos

sin ≥ 4

Proposed by Nguyen Van Nho-Vietnam IX.4. In Δ , , , - cevians, ∈ ( ), ∈ ( ), ∈ ( )

, ,Γ , Γ – inradii, respectively circumradii in Δ ,Δ , ,Γ , Γ – inradii, respectively circumradii in Δ ,Δ , ,Γ ,Γ – inradii, respectively circumradii in Δ ,Δ

Prove that: Γ ⋅ Γ ⋅ Γ ⋅ Γ ⋅ Γ ⋅ Γ ≥ 27 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

Proposed by Marian Ursărescu – Romania IX.5. Prove that in any Δ the following relationship holds:

min tan 2 , tan 2 , tan 2 ≤−+ ≤ max tan 2 , tan 2 , tan 2

Proposed by Marian Ursărescu – Romania IX.6. Let , , , > 0, with + + + = 4. Prove that:

1+

1+

1+

1+ 4

1+ +

1+ +

1+ +

1+ +

1+ +

1+ ≥

≥ 4 + 61+ 1 +

1+ 1 +

1+ 1 +

1+ 1

Proposed by Andrei Ștefan Mihalcea-Romania

IX.7. Let , , ∈ [0; +∞) ∧∑ ( , , ) = 1. Prove:

( , , )

≤ 3 + ( − 1)( , , )

Proposed by Nguyen Van Nho- Vietnam IX.8. Let , , ∈ (0; +∞) ∧ + + ≤ 3. Prove:

1 + + 1 + + 1 + ≤ 3√2 Proposed by Nguyen Van Nho- Vietnam

IX.9. Let , , ∈ [0; +∞) ∧ + + = 3. Prove:

+ + ≤ 81 + Proposed by Nguyen Van Nho- Vietnam

IX.10. Let , , ∈ [0; +∞) ∧ ∑ ( , , ) = 1. Prove:

+ 3( , , )

≤ 2√3 + 2

Proposed by Nguyen Van Nho- Vietnam IX.11. If , ∈ [0; +∞) and ∈ ℕ∗ ∧ ≥ 2 then:

( ) ≤∑

+ 1 ≤+2

Proposed by Nguyen Van Nho-Vietnam IX.12. Prove that in any Δ the following inequality holds:



tan

tan+

tan

tan+

tan

tan≥ 3 tan 2 + tan 2 + tan 2

Proposed by Marian Ursărescu – Romania IX.13. Let be Δ , the circumcenter and , , the circumradii of Δ ,Δ and

Δ . Prove that: sin + sin + sin ≤ √

Proposed by Marian Ursărescu – Romania IX.14. Prove that in any Δ the following inequality holds:

+ + ≥ 4 − 2

Proposed by Marian Ursărescu – Romania IX.15. In Δ the following relationship holds:

6≤

ℎ+ +

ℎ+ +

ℎ+

Proposed by Seyran Ibrahimov-Azerbaijan IX.16. If , > 0 then in Δ the following relationship holds:

ℎℎ ℎ ( ℎ + ℎ ) +

ℎℎ ℎ ( ℎ + ℎ ) +

ℎℎ ℎ ( ℎ + ℎ ) ≥

9( + )

Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania

IX.17. If , , ∈ 0, then in Δ the following relationship holds: (sin + tan )

sin sin+

(sin + tan )sin sin

+(sin + tan )

sin sin≥ 16√3

Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania

IX.18. In Δ the following relationship holds: ( + )ℎ + ( + )ℎ + ( + )ℎ ≥ 12

Proposed by Seyran Ibrahimov- Azerbaijan IX.19. In Δ the following relationship holds:

2 ⋅ sin ≥ (sin + sin ) + (sin + sin ) + (sin + sin )

Proposed by Seyran Ibrahimov- Azerbaijan IX.20. In Δ the following relationship holds: + + + + + ≥ 24

Proposed by Seyran Ibrahimov-Azerbaijan

IX.21. Prove that: 27

2cos414

eccos

Proposed by Vasile Mircea Popa-Romania

IX.22. Să se demonstreze egalitatea:



3 3333 1337136cos

134cos2

133cos

132cos2

135cos

13cos2


IX.23. In any Δ the following inequality holds:

1+

1+

1>

87

Proposed by Marian Ursărescu – Romania IX.24. Let , , be nonnegative real numbrs such that + + = 3. Prove:

12 + 3 +

12 + 3 +

12 + 3 ≥

6+ 9

Proposed by Le Khanh Sy-Vietnam IX.25. Prove that:

cos213 cos

313 =

√136 cos

13 cos

52√13

+1

12

Proposed by Vasile Mircea Popa – Romania IX.26. Prove that in any acute-angled triangle the following inequality holds:

min(sin sin cos , sin sin cos , sin sin sin ) ≤12 + ≤

≤ max(sin sin cos , sin sin cos , sin sin cos ) Proposed by Marian Ursărescu – Romania

IX.27. Let be an acute-angled triangle and circumcenter. We denote with , , and , , the inscribed circumradii in Δ ,Δ ,Δ . Prove that:

1+

1+

1≥

16 √3 + 2

Proposed by Marian Ursărescu – Romania IX.28. Solve for real numbers:

tan + tan = − tantan + tan = − tantan + tan = − tan

Proposed by Seyran Ibrahimov-Maasilli-Azerbaijan IX.29. If , , ∈ ℝ ∧ + + = , ∈ ℕ. Show:

+ + + + √ + ≤ 3 1 + ⋅ 3

Proposed by Seyran Ibrahimov-Maasilli-Azerbaijan IX.30. Let Δ be acute-angled and , , the points in which the heights of the triangle intersects the circumcenter. If = , then Δ is equilateral.

Proposed by Marian Ursărescu – Romania IX.31. Prove that in any Δ the following inequality holds:

( + )( + )+

( + )( + )+

( + )( + )≤ 4 + 2 + 3

Proposed by Marian Ursărescu – Romania IX.32. In any Δ the following inequality holds:



+ + ≤94 ⋅

Proposed by Marian Ursărescu – Romania IX.33. Prove that in any acute-angled Δ the following inequality holds:

max cot 2 , cot 2 , cot 2 ≥ 7 − 2

Proposed by Marian Ursărescu – Romania IX.34. In Δ the following relationship holds:

2( + + ) ≤ ( + + )(2 − max( , , )) Proposed by Seyran Ibrahimov- Azerbaijan

IX.35. Let , , be positive real numbers such that: + + = 3. Prove that:

+ + +1

+ + +1

+ + +1

+ + ≥ 4

Proposed by Hoang Le Nhat Tung –Vietnam IX.36. Let , , be positive real numbers such that: + + = 3 . Find the maximum value of:

=( + )

− + + 1 +( + )

− + + 1 +( + )

− + + 1

Proposed by Hoang Le Nhat Tung –Vietnam

IX.37. Let be the centroid in Δ and ∈ ; ∈ ; ∈ . Prove that in Δ the following relationship holds:

( + )⋅ ℎ +

( + )⋅ ℎ +

( + ) ⋅ ℎ ≥ 4√3

Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți-Romania IX.38. Let Δ be the circumcevian triangle of incentre of Δ and , > 0. Prove that:

ℎ ℎ ( ℎ + ℎ ) + ℎ ℎ ( ℎ + ℎ ) + ℎ ℎ ( ℎ + ℎ ) ≥9

( + )

Proposed by D.M. Bătinețu – Giurgiu, Claudia Nănuți-Romania IX.39. If , ∈ ℕ; ∈ ; ∈ ; ∈ then in Δ the following relationship holds:

++ℎ ⋅ + + +

+ℎ ⋅

⋅ + + ++

⋅ + ℎ ≥

≥ 4√3( + 1)( + 1) Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru-Romania

IX.40. If , , , ∈ ℕ − {0}, > > > then: (2 − 1)(2 − 1) > (2 − 1)(2 − 1)

Proposed by Daniel Sitaru – Romania IX.41. Let , , > 0. Prove that:

++

++ +

++ ≥

2 + 2 + 43 + +

2 + 4 + 23 +




IX.42. If , , ∈ 0, then in Δ the following relationshp holds: sin + tan

+ +sin + tan

+ +sin + tan

+ > 4√3

Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania IX.43. If , ∈ ℕ then in Δ the following relationship holds:

( + ( ) )( + 1) + ( + )( + ) + ( + 1)( + ( ) ) ≥ ≥ 4√3( + 1)( + 1)

Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru-Romania IX.44. Let , , > 0. Show that:

++

+≥ 2( + + 2 )

2+ 3 +

23 + −

1+


IX.45. In Δ the following relationship holds:

+ + ≥2

+2

Proposed by Bogdan Fustei-Romania IX.46. In ∆ the following relationship holds:

+ +√

≤ 2(ℎ + ℎ + ℎ )


√34 sin sin

≥(ℎ + ℎ + ℎ )( + + )

( + + ) ℎ


+ + ≥ 3 where , , circumradii of Δ ,Δ ,Δ

Proposed by Bogdan Fustei-Romania

IX.49. In ∆ the following relationship holds:

+ + ≥(ℎ + ℎ + ℎ )( + + )

+ +

where , , circumradii Δ ,Δ ,Δ . Proposed by Bogdan Fustei-Romania IX.50. In ∆ the following relationship holds:







+ +2 ≤ + +


+ +2 >

( − )( − )

Proposed by Bogdan Fustei-Romania IX.54. In Δ the following relationship holds:

3(2 + ) ≥ℎ

+ℎ

+ℎ

Proposed by Seyran Ibrahimov- Azerbaijan All solutions for proposed problems can be finded on the


10-CLASS-STANDARD

X.1. 1. If , , > 0 then in Δ the following relationship holds:

( + )ℎ +

( + )ℎ +

( + )ℎ ≥ 4√3

Proposed by D.M. Bătinețu – Giurgiu, Claudia Nănuți-Romania

X.2. If , > 0; – centroid; = ( , ); = ( , ); = ( , ) then in Δ the following relationship holds:

( + ) + ( + ) + ( + ) ≥243

( + )


X.3. In Δ , – centroid; = ( , ); = ( , ); = ( , ). Prove that:

( + ) + ( + ) + ( + ) ≥81

( + )

Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți-Romania X.4. If , , , ∈ ℕ; , , > 0 then:

( + 1)( + ) + ( + )( + ) + + ( ) ( + 1) ⋅



⋅ ( + )( + ) + ( + )( + ) + ( + 1)( + ) ≥ ≥ ( + 1)( + 1)( + 1)( + 1) ( + + )


X.5. In Δ ; – centroid; = ( , ); = ( , ); = ( , ), ∈ ; ∈ ; ∈ . Prove that:

( + ) + ( + ) + ( + ) ≥81

( + )

Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți-Romania

X.6. If ≥ 0; , , > 0; , ∈ ℕ then:

(( + ) +1

2 + 3√+ ( + ) +

12 + 3√

+

+ ( + ) +1

2 + 3√≥

3( + 1) ( + 1)5


X.7. In Δ the following relationship holds: ( + + )( + + ) ≥ 27√3 ; = [ ]


X.8. If , , ∈ 0, then in Δ the following relationship holds: sin + tansin + sin ⋅ +

sin + tansin + sin ⋅ +

sin + tansin + sin > 4√3


X.9. If , > 0;Δ – circumcevian triangle of incentre in Δ ; – centroid then in Δ the following relationship holds:

ℎ ℎ ( ℎ + ℎ ) + ℎ ℎ ( ℎ + ℎ ) + ℎ ℎ ( ℎ + ℎ ) ≥3

( + )

Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți-Romania X.10. In Δ the following relationship holds:

( + )ℎ +

( + )ℎ +

( + )ℎ ≥ 4( + 1)√3 − 3 ; ∈ ℕ

Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania

X.11. If , , ∈ 0, then in Δ the following relationship holds: (sin + tan )

sin sin +(sin + tan )

sin sin +(sin + tan )

sin sin > 16√3


X.12. In Δ the following relationship holds:



√⋅ ℎ ℎ − 2 ≥

√23 +

Proposed by Bogdan Fustei – Romania X.13 In Δ the following relationship holds:

6√3

+ +≤ +

Proposed by Seyran Ibrahimov- Azerbaijan X.14. Solve for real numbers:

tan + tan = − tantan + tan = − tantan + tan = − tan

Proposed by Seyran Ibrahimov- Azerbaijan

X.15. Solve for real numbers: ≠ ∧ ∈ ℝ, − =

=

Proposed by Urfan Aliyev-Azerbaijan X.16. Prove that in any Δ the following inequality holds:

− 2 ≥(( − ) + ( − ) + ( − ) )

3

Proposed by Adil Abdullayev- Azerbaijan X.17. Prove that in any Δ the following inequality holds:

ℎ ≥( + + + + + )(ℎ + ℎ + ℎ )∑

+ + + ℎ + ℎ − 3

Proposed by Bogdan Fustei-Romania X.18. Let ∈ ℝ. Solve in ℝ:

√1 + + √1 + 2 + √1 + 3 + √1 + 4 + √1 + 5 = 5 Proposed by Nguyen Van Nho- Vietnam

X.19. Let , , ∈ [0; +∞). Prove: (1 + )(1 + )(1 + ) ≤

Proposed by Nguyen Van Nho- Vietnam X.20. Let , , ∈ [1; 2] ∧ + + = 6. Prove: + + ≤ 18

Proposed by Nguyen Van Nho- Vietnam X.21. Let ∈ ℕ ∧ ≥ 2. Solve in ℝ:

∑ log (2− )∑ √1 +

= 1

Proposed by Nguyen Van Nho- Vietnam X.22. Let , , ∈ (0; +∞) ∧ ∑ ( , , ) = 3. Prove:



( , , )

≤ 3

Proposed by Nguyen Van Nho- Vietnam X.23. Prove that in any Δ the following inequality holds:

ℎ ≤ ℎ − 2 ≤ ℎ Proposed by Bogdan Fustei-Romania

X.24. Prove that in any Δ the following inequality holds:

≥ 31ℎ +

1ℎ +

1ℎ

Proposed by Bogdan Fustei-Romania X.25. Prove that in any Δ the following inequality holds:

+ +2 ≥ +

12 ℎ


ℎ ≥ 1 +2

Proposed by Bogdan Fustei-Romania X.27. In all Δ prove:

(4 + ) − 12> 3√3 −

32

Proposed by Rovsen Pirguliyev-Azerbaijan X.28. Prove that in any Δ the following inequality holds:

−ℎ ℎ ≥ 3√2

+ + + + ++ + + ℎ + ℎ + ℎ − 3

Proposed by Bogdan Fustei-Romania X.29. Let be any tetrahedron , , , , the exradii of the spheres and = inradii of the sphere. Prove that: a) + + +

√≤

√

b) + + + √ ≥√

Proposed by Marian Ursărescu – Romania


sin + sin + sin + sin + sin + sin ≥ 4




X.31. Prove that in any Δ the following inequality holds: + + + + +

+ + + ℎ + ℎ + ℎ − 3 ≤+ +

2∑ℎ − ∑


≥ 2 − 12


+ ℎ − 3 ≥ + ℎ ℎ ℎ

Proposed by Bogdan Fustei-Romania X.34. Let be Δ , , , excentres. Prove that:

92 ≤ + + ≤

94


X.35. Prove that in any acute-angled triangle the following inequality holds: min{( + ) cot , ( + ) cot , ( + ) cot } ≤ 2 ≤ max{( + ) cot , ( + ) cot , ( + ) cot }



min + , + , + ≤−

+ 2 ≤ max + , + , +


X.37. Let be a tetrahedron, the centroid and any point in space. Prove that:

≥23 ( ∙ )

Proposed by Marian Ursărescu – Romania X.38. Prove that in any Δ the following inequality holds:

2≥

+ +

ℎ + ℎ + ℎ


+ ++ + ≤ 2




2 ⋅ ≥ ℎ

Proposed by Bogdan Fustei-Romania X.41. Let be a tetrahedron and , , , the excenters of the sphere of the tetrahedron. (centrele sferelor exinscrise tetraedrului). Prove that:

+ + + ≥ 2 Proposed by Marian Ursărescu – Romania

X.42. Prove that in any acute-angled triangle the following relationships holds: a) ≥ 18 min cot , cot , cot

b) ≤ max cot , cot , cot Proposed by Marian Ursărescu – Romania

X.43. Prove that in any acute-angled Δ the following relationship holds:

ℎ + ℎ + ℎ ≤ 3( + ) Proposed by Marian Ursărescu – Romania

X.44. Let be Δ , – incentre and , , – circumradii of Δ ,Δ ,Δ . Prove that:

2 − 2 − ≤ 2 + 2 + 2 ≤ 4 − 8 + 3


X.45. Prove that in any Δ the following inequality holds: sin + sin

sin + sin ≤2( + )

[ ] , [ ] = Δ

Proposed by Marian Ursărescu – Romania X.46. Δ (sides , , ,),Δ (sides , , ) . Prove that:

≥23 + +


X.47. Does there exists non-constant functions : (1; +∞) → ℝ such that:

( ) = ( ) + ( ) + ⋯+ ( )? Proposed by Rovsen Pirguliyev-Azerbaijan

X.48. In ∆ the following relationship holds:

(ℎ + )ℎ

≥2

43

+83




X.49. In ∆ the following relationship holds:

+ + ≤ + +

Proposed by Bogdan Fustei-Romania X.50. In ∆ the following relationship holds:

ℎ + ℎ + ℎ ≥ 2 ⋅ℎ + ℎ

sin 2


ℎ ℎ ℎ≥

(ℎ − 2 )(ℎ − 2 )(ℎ − 2 )


+ + ≥ 2ℎ + ℎℎ +

ℎ + ℎℎ +

ℎ + ℎℎ


√2 4 + ≥ℎ

ℎ − 2


12 + + ≥ sin 2


+ + ≥ℎ ℎ

+ℎ ℎ

+ℎ ℎ


2 2 + > ℎ + ℎ + ℎ


X.57. In ∆ the following relationship holds: + + ≥ √3, , , – circumradii (razele cercurilor circumscrise), Δ ,Δ ,Δ .


X.58. Prove that in any triangle the following inequality holds:



ℎ− ≥

23√27

+ 2

Proposed by Bogdan Fustei – Romania X.59. Fie n ≥ 3 și z1 , z2 , … , zn ∈ ℂ astfel încât

|S| = |S-z1| + | S – z2 |+ | S – z3 | + … + | S – zn | , unde S = z1 +z2 + … + zn .

Proposed by Butaru Zizi, Bețiu Anicuța-Romania

X.60. Prove that in any triangle the following inequality holds:

+ + ≥ 6

Proposed by Seyran Ibrahimov –Azerbaijan

X.61. Let , , , be strictly positive real numbers such that + + + = 1. Prove that:

1 + + + 1 + + + 1 + + + 1 + + + 1 + + + 1 + + ≤14


X.62. Prove that the following is a perfect square number.

3 4 5 … 99 100

Find the smallest value of such that above number is perfect square number.

Proposed by Naren Bhandari-Nepal

X.63. In ABC the following relationship holds:

2 1 1 43 5aRp r p

b c r

.

Proposed by Marin Chirciu-Romania X.64. In ABC the following relationship holds:

222 1 13 3a

rp h pR b c

.

Proposed by Marin Chirciu-Romania X.65. In ABC the following relationship holds:

2218 1 1 9a

r hR b c

.


X.66. In ABC the following relationship holds:



22 1 1 9

2aRr

b c r

.




11-CLASS-STANDARD

XI.1. Let be : 0, → (0,∞); ( ) = tan + sin and , , ∈ 0, . Prove that in Δ the following relationship holds:

( ) ⋅ ( )sin ⋅ +

( ) ⋅ ( )sin ⋅ +

( ) ⋅ ( )sin ⋅ ≥ 16√3

Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți-Romania

XI.2. If , , > 0; < ; ∈ ℕ; ≥ 2; , ∈ (0,∞); ≤ ≤ + ; ∈ 1, ; > 0; ∈ 1, then:

+ ( + ) ≤ ( + + ) ; ∈ ℕ∗


XI.3. If , , > 0; : (0,∞) → (0,∞); ( ) =√

then: ( )

+( )

+( )

> 8√3


XI.4 Let be : (0, +∞) → ℝ, ( ) = + ln , ∈ ℕ∗ and the sequence = 1 + , ∈ ℕ∗. Find: lim → .


XI.5. Let be , ∈ (ℝ) such that 5 − √5( + ) = 3 + √5( − ) and − is invertible. Prove that: ⋮ 5. Proposed by Marian Ursărescu – Romania

XI.6. Let be > 0 and = 1 + − 1, ∈ ℕ, ≥ 2. Find: lim →



Proposed by Marian Ursărescu – Romania XI.7. Find:

lim→

1( − 1) 2 − ( − 1)(2 − 1)


lim→

arcsin−

arcsin( )

( + 1) , ∈ ℕ∗


XI.9. Let be ∈ , (ℂ) and ∈ , (ℂ) such that:

=1 1 10 1 10 0 1

Find: ( ) + ( ) . Proposed by Marian Ursărescu – Romania

XI.10.

Ω ( ) = ( − + 1)( + )!, , ∈ ℕ

Find:

Ω = lim→

Ω ( ) − ( + 1)!

Proposed by Daniel Sitaru -Romania

XI.11. Let be = 1 + + + ⋯+( )

. Find:

lim→

− 90


XI.12. Let be , ∈ (ℝ) such that: ( ) = ⋅ . Prove that:

det( + ) + det +1

≥ (det + det ) ,∀ ∈ ℝ∗


XI.13. Let be ∈ (ℝ) invertible, such that det( + ) = 0. Prove that: =

Proposed by Marian Ursărescu – Romania XI.14. Find , , ≥ 0 such that:

= + + 2 = + + 2



= + + 2 Proposed by Seyran Ibrahimov- Azerbaijan

XI.15. If 0 ≤ , , < 1 solve the system:

arcsin + arcsin = arctanarcsin + arcsin = arctanarcsin + arcsin = arctan

Proposed by Seyran Ibrahimov- Azerbaijan

XI.16. Let **:, RRgf be functions such that:

*)()1(lim Raxfxf

x, *

)()1(lim

Rbxxg

xgx

and there exists xxf

x

)(lim

, and

xxg x

x

1)(lim

. For Rt calculate the limit

x

tx

tt

xxgxgxf

222 sin

1sincos )()()(lim .

Proposed by D.M. Bătinețu – Giurgiu, N. Stanciu – Romania

XI.17. Let **:, RRgf such that: *)()1(lim Raxfxf

x, *

)()1(lim

Rbxxg

xgx

and

there is xxf

x

)(lim

,

xxg x

x

1)(lim

. For Rt , calculate:

x

tx

tt

xxgxgxf

222 cos

1cossin )()()(lim .

Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania

XI.18. Let be the sequence > 0, = + + ⋯+ , ∈ ℕ∗. Find:

Ω = lim→

∑

√


XI.19. Let be the sequence > 0 and = ,∀ ∈ ℕ, ∈ ℕ∗, ≥ 2. Find:

lim→


XI.20. Fie n

n

n nng

)1()2( 1

, Nn şi Rx . Să se calculeze xn

xn

x

nggn

222 coscos1

sinlim




XI.21. For 0nna , n

n

n nna

)1()2( 1

, ),( x , 1)( nn xb , xn

xn

xn aanxb

222 coscos1

sin)( , find

)(lim xbnn . Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania

XI.22. Find:

lim→

∑

1 ⋅ √2! √3! ⋅… ⋅ √ !


XI.23. Let be ∈ (ℝ) such that det( + ) = 0. Prove that: ∗ = 1 + det


XI.24. Find the continuous functions , , ℎ:ℝ → ℝ with the property: +2 =

( ) + ℎ( )2 ,∀ , ∈ ℝ


XI.25. Dacă ,Rx atunci şirul 2)( nn xL ,

xnxnx

n nnnxL22

2 coscos1sin !)!1()( este

convergent. Să se calculeze )(lim xLnn .


XI.26. Let be ∈ (ℝ), invertible such that: det( + ) = 0. Prove that: ∗ + det = 1 + det ⋅ ( ∗)


XI.27. Fie RyRxyx nnnnnn ,,, *11 astfel încât există *

1lim Rxxx nnn

,

şi *)(lim Rznyynn

. Să se calculeze : nnnn

yxyx 1lim .


XI.28. Dacă 1nnx , * Rxn , *Nn astfel încât există *

1 )(lim Rxxx nnn

.

Să se calculeze : n

nn

nnnxnx

11 1lim .


XI.29 Prove that if , ∈ (ℂ) such that: ( − ) = Proposed by Marian Ursărescu – Romania

Ryynn

lim



XI.30. Let be ∈ (ℤ). If det is odd, then: det( − 2 ) ≠ 0,∀ ∈ ℤ. Proposed by Marian Ursărescu – Romania

XI.31. Fie Rx . Să se calculeze:

xnnxnnx

nnnn

222 cos)1(cos)2(1sin 1lim


XI.32. Fie Rx . Să se calculeze:

xnnxnnx

nnnn

222 sin)1(sin)2(1cos 1lim

Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania XI.33. Find:

lim→

1√

arctan +12 arctan( − 1) + ⋯+

1arctan 1 ; ∈ ℕ∗, ≥ 2

Proposed by Marian Ursărescu – Romania XI.34. Let be ∈ (ℤ). Find that: det( + + ∗) + det(− + + ∗) + det(− + + ∗) + det(− ∗ + + ) ⋮ 12


=→ →

+ − 1( + + 1)!


XI.36.

( ) =→

13

(3 )

If a, b, c ∈ 0, then: 4 ( ) + ( ) + ( ) ≤ 3( + + )


XI.37.

Ω =

⎝

⎛ (2x + 3x + 1) ⋅ cos (kx) dx

⎠

⎞

Find: =

→( − ⋅ )

Proposed by Daniel Sitaru – Romania XI.38. Find:



Ω = logn + 3n + 2

n + 3n +14 log

(n− 1) + 3(n − 1) + 2(n− 1) + 3(n− 1) + ⋯+

1n log

32


XI.39. Find:

Ω = lim→

sin

√k + n


XI.40. Let be , > 0, ≠ such that lim → = lim → = , > 0. Find:

Ω = lim→

−

−

Proposed by Marian Ursărescu – Romania XI.41. Let be the sequences: ∈ (0,1), = 1 − , > 0, = + ,

∀ ∈ ℕ, ∈ ℕ, ≥ 2. Find: lim→

⋅ Proposed by Marian Ursărescu – Romania

XI.42.

( ) =2 + 2 + − 1

(2 + 2 + 2)‼

Find:

=→

! ⋅ ( )


XI.43. If > 0 then:

+ ( ) 11 + +

11 + > ( ) + ( ) 1

1 + +1

1 +


XI.44. Solve the equation: sin

[ ]+ cos

[ ]= tan

[ ] , where [⋅] denote the integer part.

Proposed by Rovsen Pirguliyev- Azerbaijan XI.45. For all > 0, prove:

sin −

ln√

+cos −

ln√

> |sin | + |cos |

Proposed by Rovsen Pirguliyev- Azerbaijan

XI.46. Prove without computer: >



Proposed by Rovsen Pirguliyev- Azerbaijan XI.47. Let be ∈ (ℝ), invertible such that det( + ) = 0. Prove that:

+ det = det ( ) ∗

Proposed by Marian Ursărescu – Romania XI.48 Let be , ∈ (ℝ) such that det = det . Prove that:

det( + ) ≥ det( + ) ,∀ , ∈ ℝ Proposed by Marian Ursărescu – Romania

XI.49. Find:

lim→

1 + 2 √2 + 3 √3 + ⋯+ √(1 + 3 + ⋯+ (2 − 1) ) ,

, ∈ ℕ∗. Proposed by Marian Ursărescu – Romania XI.50. Let be , > 0, = − , = + , ∈ ℕ∗, ≥ 2. Find:

lim→

( ) Proposed by Marian Ursărescu – Romania

XI.51. Ω (x) = ∫ dt, n ∈ ℕ, n ≥ 1. Find:

=→

( − )− ( )( − 1)

Proposed by Ahmad Albaw-Jordan XI.52. Find:

=→

1 −1 + √

1 + √


XI.53. x = 1, x = 3, x = x + 2x , n ≥ 3. Find:

=→

(−1) ( − 2 − )

Proposed by Daniel Sitaru – Romania XI.54. Let be the sequences: > 0, = ln(1 + ) ,∀∈ ℕ and > 0,

= ⋯ . Find: lim → ( ). Proposed by Marian Ursărescu – Romania

XI.55.

α(x) =43 −

13 ⋅ cos (3 x) , x ∈ 0,

π2 , β(x) = α

π2 − x

Find:

=→ →

( ) (3 ) ⋅ … ⋅ (2 − 1)(2 ) (4 ) ⋅ … ⋅ (2 )



Proposed by Daniel Sitaru – Romania XI.56. , , > 0,2 ( + + ) = 3 + 2,Ω( ) = lim → − .Prove that:

Ω( ) ⋅ Ω( ) ⋅ Ω( ) ≤1

27

Proposed by Daniel Sitaru – Romania XI.57. Find:

=1

3 3 ⋅3

− − 1


XI.58 Ω( ) = − + 4∑( )( )( )

, ∈ ℝ

If ∈ (0,1), > 1 then: Ω( ) ( )+ Ω( ) ( )

< 1 + Ω( ) ⋅ Ω( )




12-CLASS-STANDARD

XII.1. Find:

lnsin(arctan )

√1 +

Proposed by Abdul Mukhtar-Nigeria XII.2. Find:

cos + sinsin(2 )− sin (2 )

Proposed by Abdul Mukhtar-Nigeria

XII.3. Find:



2 + 5 + 6 + 6 + 12( + 2 + 2)√ + 2 + 2

Proposed by Abdul Mukhtar-Nigeria XII.4. Find:

(4 + 2 + )+

ln(1 − )2 − 2 + 1


XII.5. Let ∈ ℕ∗ and ∈ ℝ,∀ = 1; . Find Ω = ∫ ln(∏ ( − )) .

> max |∀ ∈ 1; Proposed by Nguyen Van Nho-Vietnam

XII.6. Let be :ℝ → ℝ continuous such that ( ) + ( ) + ( ) = − ,∀ ∈ ℝ, >0. Find: ∫ ( ) . Proposed by Marian Ursărescu – Romania XII.7. Find:

+ 2 + 3 + 4 + 5− 7 + 5 + 5 + 9 + 52 − 60

Proposed by Seyran Ibrahimov- Azerbaijan XII.8. Find:

=+ 3 + 5 + 7

7 + 14 − 2 − + 2 − 11 − 6 , ∈ ℝ


Ω =+ + +

+ + + + + + + , > 0

Proposed by Naren Bhandari-Nepal XII.10. Find:

=2 + 3 + 9 + 4+ 2 + 9 + 8 + 16


+ + + 1+ 10 + 28 + 28 + 27 + 18 , > 0




=→

1 31 + 3 −

21 + 2

Proposed by Daniel Sitaru– Romania XII.13. Find:

=→

∫ √ ⋅ √

∫ √ ⋅ √


=→

2 + 3( + 1)( + 2)( + 3) +


=→∞

( + 1)

⎝

⎛+

⎠

⎞

Proposed by Nho Nguyen Van-Vietnam XII.16. Find:

=→

1( + 1)

++


=→

1( + 1)(2 + 1)

( + 1)(2 + 1)( + 1)(2 + 1)


=→

12 +

32 +

32 +

3

Proposed by Daniel Sitaru – Romania XII.19.

( ) =(1 + 2 + ) (1 − 2 + )

(1 − )(1 + 2 + ) , ∈ ℕ, ≥ 2

Find:



=→

( + 2) − ( + 1)( + 1)− ( )


XII.20. Find: Ω = lim →∫

∫

Proposed by Daniel Sitaru– Romania XII.21. Let : [0,1] → (−1,1) be a continuous function so that ∫ ( ) ∉ {−1,1}.Prove

that: ∫ ( )

∫ ( )≤ ∫

( )

( ) .

Proposed by Soumitra Mandal - India XII.22. If 0 < < < then:

7 tan sin 2 + tan 2 √sin tan > −


XII.23. Let f:ℝ → ℝ be integrable and satisfying

f(xt + (1 − t)y) ≤ tf(x) + (1 − t)f(y) where x, y ∈ ℝ and t ∈ (0,1) Show that:

1ln a f(x ) dx +

1ln b f(x ) dx ≥ 2 (ab) f(a − a b + a b − a b + b )dx

where , > 0. Proposed by Soumitra Mandal-India

XII.24. If 0 < < < then:

cossin + 4 cos <

15 log ⋅ log

sinsin


XII.25 Find: Ω = lim → ∫ nx + − nx dx, [∗] - great integer function


XII.26. Find:

Ω = lim→

tan√1 + x − 1

x dx




XII.27. If 0 < < 1, : , → ℝ, convexe and increasing function then:

1− 1 +2 ≤

1 +2

( ) ≤1 −

2( ) +

1

Proposed by Abdallah El Farissi-Bechar-Algerie

XII.28. If , , ∈ 0, then:

0 ≤ (tan + 2 tan tan ) + 4 tan ≤

Proposed by Daniel Sitaru – Romania XII.29. If , , > 0 then:

∑ ∫ ≤ ⋅ ⋅


XII.30. If , , ≥ 0, ( ) = ∫ ( )( )( )

then:

( ) + ( ) + ( ) ≥ log + 1 +



Magazine-Interactive Journal. UNDERGRADUATE PROBLEMS

U.1. An intriguing mock exponential integral with an improper closed-form.

( − 1)( + 1) ⋅

tanh ( )= −

4− 3 log √2 − log(7) +



+5(log(3)− log(5)) + logΓ Γ

Γ Γ

– Catalan Constant. Proposed by Srinivasa Raghava-AIRMC-India

U.2. Application of the Fourier series:

sin (2 + 1)+ (2 + 1) (−1) = 4 ⋅

sinh( )

cosh

provided | | ≤ & > 0, > 0

sin (2 + 1)

+ (2 + 1) (−1) = 2 ⋅tanh

Proposed by Srinivasa Raghava-AIRMC-India U.3. Find:

Ω =− 2 log

+ log (1 + )

Proposed by Khalef Ruhemi-Jarash-Jordan U.4. Find:

ln + √ + 1√ + 1

Proposed by Abdul Mukhtar-Nigeria U.5. Evaluate the following integral:

sin sin( ) sin(2 )

and prove that

sin sin( ) sin(2 ) ≈1320

Proposed by K. Srinivasa Raghava – AIRMC – India U.6. If we define the function ( ) for any > 1

( ) = log ( ) (tanh ( )− tanh ( ))√

Then prove that:

lim→

( ) = 14 (3)−12

( ( − 4) + 8) + 8 log(2) − 16 − 8

Proposed by K. Srinivasa Raghava – AIRMC – India U.7. Find:

Ω =⋅ tan+ + 1



Proposed by Vasile Mircea Popa – Romania U.8. Prove that:

2 +sin( )sinh(2 ) ( coth( )− 1) =

23 +

536

Proposed by Srinivasa Raghava-AIRMC-India U.9. If

(coth( ) + cos( )) = ( ) + (tanh( ) + sin( ))

then we have

√ + −√√

= 2( − 1)

Proposed by Srinivasa Raghava-AIRMC-India U.10. For sufficiently large positive integer (some times we can say approaches to infinity), we have:

1( − 1) ≥ log( ) + 2

here = .57721566 … Euler’s constant Proposed by Srinivasa Raghava-AIRMC-India

U.11. Show that:

1√2

+ sin 2 + tanh 2 √=

= 2 1 + √2 − 112 +

√17 − 417

Proposed by K. Srinivasa Raghava – AIRMC – India U.12. Evaluate the integral:

ln 1 + − ln 1 − ln( )

and then prove this sharp inequality

ln 1 + − ln 1 − ln( ) ≤ 4

Proposed by Srinivasa Raghava-AIRMC-India U.13. Prove that:

(tan ( ) + tan ( )) log(tan( )) log(tan( )) (tan( ) tan( ))tan( ) tan( )

= ( ) − (6)− , Where is the Catalan’s constant, ( ) denotes the Polygamma function, designates the

Riemann zeta function and ( ) represents the inverse tangent integral. Proposed by Cornel Ioan Vălean – Romania



U.14. Evaluate the sum in closed form, for ≠ 0 1

sinh + (2 + 1), ∈

Proposed by Srinivasa Raghava-AIRMC-India U.15. For any ≥ 1, we have

2 1 − cos( ) + log(cos( )) + log(cos ( )) =

= ( − ( + 1) log(2))−√ Γ

Γ 1 +

2 1 − cos( ) + log(cos( )) + log (cos( )) =

= −√ Γ

Γ 1 ++

12 log(2) (log(2) − 2 ) + 24

Inspired from Kartick Chandra Betal Proposed by K. Srinivasa Raghava – AIRMC – India

U.16. Prove that: 1

5 + 4 +1

5 + 11

5 + 3 +1

5 + 2 (−1)4 + 1 ≥

where is Euler’s constant. Proposed by K. Srinivasa Raghava – AIRMC – India

U.17. Calculate the integral:

− + 1

It is required to express the integral value with the usual mathematical constants, without using values of special functions. Proposed by Vasile Mircea Popa-Romania

U.18. Find:

Ω = lim→

|sin |


U.19. Prove that:

sin( )sin( )

ℭ

(cos ( )− cos ( )) =sin ( )

sin( )

ℭ

(cos ( )− cos ( ))

where {ℭ, , , , } are integers, ℭ < { , } , ≥ℭ

,ℭ



Proposed by Ahmed Albaw-India U.20. Calculate the integral:

ln √ √√

2 +

Proposed by Ekpo Samuel-Nigeria U.21. Calculate the integral:

log( ) sin ( )

Proposed by Zaharia Burghelea-Romania U.22. Find:

Ω = lim→

1 +1 −

Proposed by Ekpo Samuel-Nigeria U.23.

! + ≤ ≤ ! + Is there any prime number such that given inequality above can be true, where 1 ≤ ≤ and > 2?

Proposed by Naren Bhandari-Nepal U.24. Prove that for ∈ ℤ

…

= √…

Setting = = = ⋯ = = 2 for = 5 also prove that:

2√

= √2√

√√

√

< 2 Proposed by Naren Bhandari-Nepal

U.25.

cos 128 =

⎝

⎜⎜⎜⎜⎛ 1 + √2 + 2√2 + 2 2√2 + 2 2 2√2 + + 2 2 2 2√2

2 2 2 2 2√2⎠

⎟⎟⎟⎟⎞

Proposed by Naren Bhandari-Nepal



U.26. ( ) − 1 ( ) + 3 ( ) ( ) + ( )

+2 (2 + 1)(2 + 1) =

= 60 − , , ∈ ℝ, ( ) – sine integral function Find:

Ω = + +(−1)

cos 4

Proposed by Arafat Rahman Akib-Bangladesh U.27.

√+

1√ √

= ⋅ ( )

Find: Ω = ℒ ( ) ⋅ log √2 ,ℒ - Lapalce’s transform

Proposed by Arafat Rahman Akib-Bangladesh U.28. Let

( ) = 1 + ( − 1) + 1+ 1

(−1)+ 1

then for any ≥ 1, ≥ 1, we have

( ) = ( ) −

Note: ( ) = – Harmonic Number

Proposed by Srinivasa Raghava-AIRMC-India U.29. An Unique Integral

+ +(log( ) + log( ) + log( )) = 3

Proposed by Srinivasa Raghava-AIRMC-India U.30. If,

Φ( ) =cosh( ) ( )

(2 )!

then

( ) = 2 ( ) ln( )

where = √−1 Proposed by Srinivasa Raghava-AIRMC-India

U.31. ( , , ) = ∫ sin ; , > 0, = −2,−1,0,1 … Find:



Ω = lim→

lim→

lim→

( , , )

√

Proposed by Feti Sinani-Kosovo U.32. Find:

→

( )

( ) 1 +

( )

( )

−

1 + −

where ( ) is Riemann Zeta function and Γ( ) is Gamma function. Proposed by Feti Sinani-Kosovo

U.33. Calculate the integrals:

ln(1 + tan + tan ) ,1

ln(1 + tan + tan ) , tan√1 + − 1

Proposed by Abdul Mukhtar-Nigeria All solutions for proposed problems can be finded on the


ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-sUMMER 2018

PROBLEMS FOR JUNIORS

JP.136. Let , , be positive real numbers such that: = . Find the maximum of the

expression:

=+ − +

++ − +

++ − +

Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam

JP.137. Let , ≥ . Prove that:



+ ≥ +( − )

( + + )( + + )

Proposed by Andrei Ștefan Mihalcea – Romania

JP.138. Let , , > 0, with + + = . Prove that:

( − )( − )( − ) ≥ √


JP.139. Let , , be positive real numbers such that: + + = . Find the minimum

of the expression:

=+

++

++


JP.140. Let , , > 0. Prove that:

√ +≤ −

∑∑



+≤

(∑ )


JP.142. Let , , ≥ . Prove that:

−≤ −

∑


JP.143. In any triangle the following relationship holds:

⋅ + ⋅ + ⋅ ≤ −

all notations are usual sense.

Proposed by Mehmet Șahin – Ankara – Turkey




+ + ≥



+ + + + + ≤


JP.146. Let , , be positive real numbers such that: = . Find the maximum of the

expression:

=( − + )

+( − + )

+( − + )


JP.147. Let , , be positive real numbers such that: + + = . Find the

minimum of the expression:

=( + )

+( + )

+( + )


JP.148. Let , , be positive real numbers such that: + + = . Prove that:

+− + +

+− + +

+− + ≥


JP.149. Find all functions: : ( , +∞) → ℝ which verify the relationship:

( ) ≤ ( ) + ( ) ≤ ( ),∀ , > 0


JP.150. Let be , , ∈ ℂ∗ different in pairs such that | | = | | = | |. If ( +

. are the affixes of an equilateral triangleࢠ ,ࢠ,ࢠ =, thenࢠࢠࢠ+ࢠ+ࢠࢠ+ࢠࢠ


PROBLEMS FOR SENIORS

SP.136. Let , , be positive real numbers such that: + + = + + . Find

the maximum of the expression:



=+

++

++


SP.137. Let , , > 0 such that: + + = . Prove that:

( + ) ++

( + ) ++

( + ) ++√ + √ + √

≥


SP.138. Let , , be positive real numbers such that: + + = . Prove that:

( + )+

( + )+

( + )≥


SP.139. In triangle the lengths of sides , , are , , . Let , , be the

distances from , , to , , ; , , are the lengths of the bisectors , , .

Prove that:

+ + ≥ + +


SP.140. Let , , be positive real numbers. Prove that:

++

++

+≥

( + + )+ + +

( + + )+ +


SP.141. Let , , > 0 such that: + + = . Prove that:

− √ ++

− √ ++

− √ +≥

+ +


SP.142. Let , , be positive real numbers such that: = . Prove that:

− + + + − + + + − + + ≤+ + +


SP.143. Let , , be non-negative real numbers. Prove that:

+ + + + + ≥ + + + + +

Proposed by Do Quoc Chinh – Ho Chi Minh – Vietnam



SP.144. Let , , be the corners in a triangle . Prove that:

+ + ≥


SP.145. If < ≤ then:

+≤

++

( )


SP.146. Let be , ∈ (ℝ) such that:

= −−

Find: (( ) − ).


SP.147. Find all continuous functions :ℝ → ℝ having the property:

( ) + ( ) + ( ) = + + ,∀ ∈ ℝ.


SP.148. Let be > 0 and = ,∀ ∈ ℕ. Find: → ⋅ .


SP.149. Let be the sequence ( ) ∈ℕ: > 1 and = + ,∀ ∈ ℕ. Find

→ .


SP.150. Let be ∈ ℤ, = + + ⋯+ + , such that

, , … , ∈ {± , ± , … , ± }. If is a prime number, > then is irreducible

over ℤ.


UNDERGRADUATE PROBLEMS

UP.136. Prove that:



( ) =14

2 + ( )

√1 −

where, ( ) and ( ) denotes the Chebyshev Polynomials of first and second kind.

Proposed by Shivam Sharma – New Delhi – India

UP.137. Let , :ℝ∗ → ℝ∗ be functions such that:

→ ( + ) − ( ) = ∈ ℝ∗ , →( )

( )= ∈ ℝ∗ and exists →

( ) and

→( ) . For ∈ ℝ calculate the limit:

→( ) ( ) − ( )


UP.138. Let , :ℝ∗ → ℝ∗ such that:

→ ( + )− ( ) = ∈ ℝ∗ , →( )

( )= ∈ ℝ∗ and exists

→( ) , →

( ) .

For ∈ ℝ, calculate:

→( ) ( ) − ( )


UP.139. Calculate:

→( + ) − ( + )

where ∈ ℝ and is the Gamma function (Euler integral of the second kind).


UP.140. Calculate:

→( + ) − ( ( + )

where ∈ ℝ and is the Gamma function (Euler integral of the second kind).




UP.141. For { } , = ( )( )

, ∈ (−∞,∞), { ( )} ,

( ) = − , find → ( )


UP.142. Let ( ) be a sequence which satisfy:

− ( + ) + =

where is positive integer and is Euler-Mascheroni’s constant. Compute: →


UP.143. Let , ∈ ℝ , ( , ) = − ( + ) + ∑ with → ( , ) =

( , ) ∈ ℝ. Calculate:

→ + + + − ( , ) .


UP.144. Let ∈ ( ,∞) and ( ) = + , for any ∈ ( ,∞). Calculate:

→ →( + )− ( ) .


UP.145. Let be ( ) , ∈ ℝ∗ ,∀ ∈ ℕ∗, such that exists → ( − ) = ∈ ℝ∗ .

Find:

→

( + )

( + )‼−

( − )‼


UP.146. Let : ( ,∞) → ( ,∞) be a function with →( ) = ∈ ( ,∞) and ∈ ℝ. Find:

→( + ) ⋅ ( ) ( ) … ( ) ( + ) − ⋅ ( ) ( ) … ( )


UP.147. In an triangle let be , , the lengths of , , , and , , exradii.

Prove that:



+ + ≥( + + )


UP.148. Let , , be positive real numbers such that: + + = . Prove that:

( + + ) + ≥ ( + + )


UP.149. Prove that:

(− )+ + + =

( + + )! ( )! ( )! ( )!( + )! ( + )! ( + )! ( )! ( )! ( )!

Proposed by Shivam Sharma – New Delhi – India

UP.150. If , , > 0 then:

+ ( )≥ ( + + )



Magazine-Interactive Journal. ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-WINTER 2018

PROBLEMS FOR JUNIORS

JP.151. Let , be positive real numbers. Find the maximum of such that inequality is

true:

+ + + ≥+

( + )




JP.152. Let ; , , denote the lengths of the highways from the peaks

, , ; , , are the lengths of the symmetric divergence lines from the peaks

, , ; , , are the radius of the circle next to the corners , , . Prove that:

+ + ≥


JP.153. In the lengths , , are , , . Let , , are the length of the

bisectors from the vertices , , in triangle . Prove that:

+ + ≤ ( + + )


JP.154. In triangle with sides = , = , = . , , are exradii,

, , are distances from , , to , , . Prove that:

+ + ≥ + +


JP.155. Let , , be positive real numbers such that: + + = . Prove that:

( + )+

( + )+

( + )≥


JP.156. Let be a triangle having the area . Let be ∈ such that the incircles of

, ′ have the same radius. Analogous, we obtain the points ∈ , ∈ .

Prove that:

=⋅ ⋅


JP.157. Prove that in , the following inequality holds:

⋅ ⋅ ⋅ ≥ ( − )





+ + ≥ +


JP.159. Prove that in any the following inequality holds:

+ + ≤ ( + )


JP.160. Prove that for all non-negative real numbers , , the following inequality holds:

+( + ) +

+( + ) +

+( + ) ≥ + + + + +

Proposed by Nguyen Viet Hung – Hanoi – Vietnam

JP.161. Let , , , be positive real numbers such that ≥ . Prove that:

+ + ++ + + +

+ + ++ + + +

+ + ++ + + +

++ + ++ + + ≤


JP.162. Let , , be non-negative real numbers such that + + = . Prove that:

(a) √

≤ +√

+ ≤ .

(b) ≤ +√

+ ≤ . When do the equalities occur?


JP.163. Let be an acute triangle with standard notations. Prove that:

+ + ≤+ +


JP.164. If , ≥ then: + + √ + ≥ √


JP.165. If , , ≥ then:

( + + ) ≤ √ − + + + + +


PROBLEMS FOR SENIORS



SP.151. Let , , be positive real numbers such that: + + = . Prove that:

( + ) ++

( + ) ++

( + ) +≥

Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam SP.152. Let , , be positive real numbers. Find the minimum of value:

=( + )

+( + )

+( + )

+ + +

Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam SP.153. Solve the system of equations:

⎩⎪⎨

⎪⎧ + = ( + ) +

− + =+


SP.154. Let , , be positive real numbers such that: + + = . Find the minimum of

value:

=√ + + √ +

√ + + √+ ( + + )


SP.155. Let , , be positive real numbers such that: = . Find the minimum of

value:

= ( − + ) + ( − + ) + ( − + ) ++ +


SP.156. Find all the polynomials ∈ ℝ[ ] having the property ( ) = + √ + ,

∀ ∈ ℝ.


SP.157. Let : [ , +∞) → [ , +∞) a derivable function and > 1. If

( )( ( ) + + + ) = ,∀ ≥ then: → ( ) exists and it is finite.

Proposed by Marian Ursărescu – Romania SP.158. Prove that for any real numbers , , … , , the inequality holds:

( + ) + … + ≥



where is a positive integer and = . When does the equality occur?


SP.159. Prove that in any triangle with standard notations, the inequality holds:

( + + )( + + ) ≥ ( + ) + ( + ) + ( + )


SP.160. Let , , be positive real numbers such that + + ≥ + + . Prove that:

( + + ) ≥ ( + + )


SP.161. Prove that the following inequalities hold for all real numbers , , ∈ [ , ]

(a) ( − + )( − + )( − + ) ≤ ( − + ).

(b) ( − + ) ( − + ) ( − + ) ≤

≤ ( − + )( − + )( − + )

When does the equality occur?


SP.162. If ∈ [ ,∞), , , , ∈ ( ,∞), then in any triangle , with usual notations

holds:

++

≥( + )

( + ) √


SP.163. If ∈ [ ,∞), , , , ∈ ( ,∞), then in any triangle , with usual notations

holds:

( + )

+≥

( + )

( + )


SP.164. If , , > 0 then:

( + ) + − + ( + ) + − + ( + ) + − ≥

≥ ( + + )


SP.165. If , , ≥ then:

( + ) + + ( + ) + + ( + ) + ≥



≥ √ − ( + + )


UNDERGRADUATE PROBLEMS

UP.151. Given real numbers , , … , ∈ [ , ]. Find the maximum and minimum

possible value of

= + + ⋯+ + ( − )( − ) … ( − )


UP.152. Let , , … , ( ≥ ) be positive real numbers such that:

≥( − )

Prove that:

+ −≤


UP.153. Find:

= ( )

Proposed by Ekpo Samuel – Nigeria

UP.154. If ∈ ( ,∞); ∈ ℕ; ≥ ; → = > 0 then find:

=→

⋅ √ ! + ( + )!

[∗] - great integer function.

Proposed by D.M. Bătinețu – Giurgiu, Daniel Sitaru – Romania

UP.155. If ∈ ℕ; , , > 0 then in the following relationship holds:

+ + + ≥ ( + )√


UP.156. If , , > 0; < ; ( + ) ∈ ℕ; ( + ) > 1; ∈ ℕ; ≥ ; ∈ [ , ]; >

0; ∈ , then:



( )

≤+ +

+ ( + ) ⋅


UP.157. If ≥ ; , > 0 then in the following relationship holds:

( + ) + ( + ) + ( + ) ≥

≥( + ) ⋅ ⋅


UP.158. Find:

= ( )

Proposed by Arafat Rahman Akib – Dhaka – Bangladesh

UP.159. Find:

= ( + + ) ( − )


UP.160. Let , , ∈ ( ; +∞) and ≥ . Prove:

([ ] + { } + ) + ([ ] + { } + ) + ([ ] + { } + ) ≥ ( + + + )

[ ] is the integer part of the real number ;

{ } is the fraction of real numbers .

Proposed by Nguyen Van Nho – Nghe An – Vietnam

UP.161. Find:

= ( + + )( + )


UP.162. If tetrahedron; = ; = ; = ; = ; = ; =

then:



+ ≥ √ [ ]

[ ] – total area of tetrahedron .


UP.163. Let , , be positive real numbers such that:

( + )( + )( + ) = . Find the minimum value of:

= ( + + )( + ) + ( + + )( + ) + ( + + )( + )


UP.164. Solve the equation: ( + ) + √ − = − Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam

UP.165. Solve the system of equations:

⎩⎨

⎧√

+√

+√

=

+ + =( + + )

Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam All solutions for proposed problems can be finded on the




INDEX OF AUTHORS RMM-22

Nr.crt. Numele și prenumele Nr.crt. Numele și prenumele 1 DANIEL SITARU-ROMANIA 28 SRINIVASA RAGHAVA-INDIA 2 D.M.BĂTINEȚU-GIURGIU-ROMANIA 29 NGUYEN VAN NHO-VIETNAM 3 CLAUDIA NĂNUȚI-ROMANIA 30 HUNG NGUYEN VIET-VIETNAM 4 NECULAI STANCIU-ROMANIA 31 PHAM QUOC SANG-VIETNAM 5 MARIAN URSĂRESCU-ROMANIA 32 ARAFAT RAHMAN AKIB-BANGLADESH 6 BOGDAN FUSTEI-ROMANIA 33 NGUYEN NGOC TU-VIETNAM 7 DAN NĂNUȚI-ROMANIA 34 GEORGE APOSTOLOPOULOS-GREECE 8 MARIN CHIRCIU-ROMANIA 35 HOANG LE NHAT TUNG-VIETNAM 9 TITU ZVONARU-ROMANIA 36 MEHMET SAHIN-TURKEY

10 TUTESCU LUCIAN-ROMANIA 37 ROVSEN PIRGULIEV-AZERBAIJAN 11 ANDREI ȘTEFAN MIHALCEA-ROMANIA 38 SHIVAM SHARMA-INDIA 12 VASILE MIRCEA POPA-ROMANIA 39 ADIL ABDULLAYEV-AZERBAIJAN 13 CREMENEANU LUIZA-ROMANIA 40 DO HUU DUC THINH-VIETNAM 14 MIHALY BENCZE-ROMANIA 41 NGUYEN PHUC TANG-VIETNAM 15 PRUNARU CONSTANTINA-ROMANIA 42 MARTIN LUKAREVSKI-MACEDONIA 16 DAN NEDEIANU-ROMANIA 43 MICHEL BATAILLE-FRANCE 17 GHEORGHE ALEXE-ROMANIA 44 NGUYEN NGOC TU-VIETNAM 18 SERBAN GEORGE FLORIN-ROMANIA 45 SEYRAN IBRAHIMOV-AZERBAIJAN 19 BUTARU ZIZI-ROMANIA 46 LE KHANH SY-VIETNAM 20 BETIU ANICUTA-ROMANIA 47 NAREN BHANDARI-NEPAL 21 IOAN CORNEL VALEAN-ROMANIA 48 URFAN ALIYEV-AZERBAIJAN 22 ZAHARIA BURGHELEA-ROMANIA 49 FETI SINANI-KOSOVO 23 IONICA CONSTANTIN-ROMANIA 50 ABDUL MUKHTAR-NIGERIA 24 CARMEN CHIRFOT-ROMANIA 51 ABDALLAH EL FARISSI-ALGERIE 25 DO QUOC CHINH-VIETNAM 52 EKPO SAMUEL-NIGERIA 26 KHALEF RUHEMI-JORDAN 53 SOUMITRA MANDAL-INDIA 27 AHMED ALBAW-JORDAN

NOTĂ: Pentru a publica probleme propuse, articole și note matematice în RMM puteți trimite materialele pe mailul: [email protected] All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.

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