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ROMANIAN MATHEMATICAL MAGAZINE

Founding EditorDANIEL SITARU

Available onlinewww.ssmrmh.ro

ISSN-L 2501-0099

RMM - Calculus Marathon 201 - 300

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RMM

CALCULUS

MARATHON

201 – 300

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Proposed by Daniel Sitaru – Romania

Marian Ursărescu – Romania

Nguyen Van Nho-Nghe An-Vietnam

Rovsen Pirguliyev-Sumgait-Azerbaidian

Nawar Alasadi-Babylon-Iraq

Mihály Bencze-Romania

Kays Tomy-Nador-Tunisia

Nishant Kumar-Jamsedpur-India

George Apostolopoulos-Messolonghi-Greece

Arafat Rahman Akib-Bangladesh

Shafiqur Rahman-Bangladesh

Uche Eliezer Okeke-Anambra-Nigeria

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Solutions by Daniel Sitaru – Romania

Chris Kyriazis-Athens-Greece, Henry Ricardo-New York-USA

Ravi Prakash-New Delhi-India, Serban George Florin-Romania

Soumitra Mandal-Chandar Nagore-India, Rozeta Atanasova-Skopje

Santos Martins Junior-Brussels-Belgium, Togrul Ehmedov-Baku-Azerbaidian

Soumava Chakraborty-Kolkata-India, Abdallah Almalih-Damascus-Syria

Rajeev Rastogi-India, Athina Kalampoka-Greece

Abdallah El Farissi-Bechar-Algerie, Khanh Hung Vu-Ho Chi Minh-Vietnam

Bedri Hajrizi-Mitrovica-Kosovo, Yen Tung Chung-Taichung-Taiwan

Nawar Alasadi-Babylon-Iraq, Nitin Gurbani-India

Igor Soposki-Skopje, Shivam Sharma-New Delhi-India

Seyran Ibrahimov-Maasilli-Azerbaidian, Dimitris Kastriotis-Athens-Greece

Lazaros Zachariadis-Thessaloniki-Greece,

Myagmarsuren Yadamsuren-Darkhan-Mongolia, Manish Moryani-India

Sanong Huayrerai-Nakon Pathom-Thailand, Radu Butelca-Romania,

Rajsekhar Azaad-Gaya-India

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201.

If ( ), ( ), ( ), , , ∈ ℂ∗, | | = | | = | | = ,∑ | | =

then:

= =

Proposed by Marian Ursarescu-Romania

Solution by Radu Butelca-Romania

We know that + + = ⇔ + = −

⇔ | + | = | − | ⇔ | + | = (1)

= + | + | =( )

+ = ( + ) =

= + ⇒ + =

| | = | | = | | = ⇒ =⇒ + =

= + ≥ + ∑ ⇔ + ≥ ⇔

⇔ + ∑ ≥ ⇔ ∑ ≥ ⇔ ∑ ≥ (2)

: ( , ), ( ) = it’s a concave function

∑ ≥ ∑ = ° = ⇒ ∑ ≥( )

⇒ ∑ = , equality holds when

is equilateral

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202.

If , , ∈ ℂ∗, | | = | | = | | = , ( + )( + )( + ) ≠ ,

( ), ( ), ( ),∑ ( )( ) = then = =


Solution by Chris Kyriazis-Athens-Greece

It’s easy to see that: ( )( )

= −( )( )

. So,

∑( )( )

= ⇔ −( + ) + ⋅ ( )( )( )( )

= (1)

But + + = + + =

So, (1) ⇒ + + = and (equivalent); + + =

We have ( + ) + = ⇔ − + = ⇒ = ⇔

− = − ⇒ ( − )( + ) = ( − ) ⇒ |− | =

⇒ | − | ⋅ + = | | ⋅ | − | ⇒ | | =

| − | = | − | ⇔ ( ) = ( ). Working just the same

= ⇔ − = − ⇔ ( − )( + ) = ( − ) ⇒ |− | =

| − | ⋅ + = | | ⋅ | − | ⇒ | | =

| − | = | − | ⇔ ( ) = ( ). So, ( ) = ( ) = ( )

203. For ∈ ∗ ∖ { } ∧ , , … , > 3. Prove:

∑ ( − )≥

Proposed by Nguyen Van Nho-Nghe An-Vietnam

Solution by Ravi Prakash-New Delhi-India

For ≥ , let ( ) = − = + −

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( ) = − + − − ( − ) = − +−

( − ) > 0∀ > 3

⇒ ( ) increases on [ ,∞) ∴ ( ) ≥ ( ) = ∀ ≥ ⇒ ≥ ∀ ≥

∴ For , , … , ∈ ℕ, ≥ , ≥ ⇒ ∑ ≥ (∑ ) ⇒

⇒∑ ( − )

≥

204.

∈ (ℝ), ( + ) = ( + + ) =

Find: =


Solution 1 by Serban George Florin-Romania

( ) = ( − ) = + + + +

+ √ ⋅ − √ = ⇒ √ =

√ + √ + √ + √ + =

− √ − + √ + = , − + + √ ( − ) =

⇒ − = , − + =

+ + = , = − = − , = = − + , = − −

( (− + ) ) ⋅ ( − (− − ) ) =

⇒ (− + ) = (− + ) + (− + ) + (− + ) + (− + ) + =

( − + ) + (− + ⋅ + (− ) ⋅ + ) + ( − + ) + (− + ) + =

− + (− + + − ) − − + + =

− + + − − + + =

( − + − ) + ( − + ) =

=− = −− + =− + =

⇒− = −− + =− + =

− = − , = −

− + =− =

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− + − =

− = − + = | ⋅− =

⇒ = ⋅ , =

= , = , = − = ⋅ = = , =

= = ( ) = =

Solution 2 by Ravi Prakash-New Delhi-India

Suppose ∈ (ℝ).

Let ( ) = ( − ) = + + + + (1) ( , , , ∈ ℝ)

be characteristic equation of .

= ( ) = ( ) = product of eigenvalues of .

As ( + ) = , we get

− √ + √ = ⇒ − √ − √ =

⇒ −√ − √ = ⇒ −√ = ⇒ −√ =

⇒ √ is an eignvalue of ⇒ √ ,−√ are zeros of ( ).

Next, ( + + ) = ⇒ {( + ) + } =

⇒ ( + ) − ( + ) + = ⇒ ( + ( − ) ) ( + ( − ) ) =

⇒ | ( + ( − ) )| = ⇒ ( + ( − ) ) =

⇒ − + is an eigenvalue of A ∴ − + ,− − are zeros of ( )

Thus, ( ) = = √ −√ (− + )(− − ) = ( )( + ) =

205.

If ∈ (ℝ) then:

( + ) = ⇔ = and ∗ =



Let = characteritstic equation of is

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( ) = ( − ) =−

−−

= ( − )( − )( − ) +

+ + + − ( − ) − ( − ) − ( − )

= − + ( + + ) − ( + + − − − )

+ ( ) = − + ( ) − [ + + ] + ( )

( + ) =

⇒ ( + )( − ) = ⇒ ( + ) ( + ) = ⇒ | ( + )| = ⇒

⇒ ( + ) = ⇒ − is an eigen value of . Thus,

−(− ) + ( ) (− ) + ( + + ) + ( ) =

⇒ ( + + − ) − ( ) + ( ) = .

Equating real and imaginary parts, we get ( ) = ( ) and ( ) = .

206. Find:

=→∞

( + )

⎝

⎛+

⎠

⎞

Proposed by Nho Nguyen Van-Nghe An-Vietnam Solution by Daniel Sitaru-Romania

= + , = + , + = =

= −−

− + −= + =

= → = , = ( + ) ∙ = ( + ) ∙

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→= < 1 →

→=

207. Prove if < < then:

−( − ) √ +

≥

Proposed by Nho Nguyen Van-Nghe An-Vietnam

Solution 1 by Soumitra Mandal-Chandar Nagore-India

Let ( ) = for all ∈ [ , ] then ( ) = > 0 for all ∈ [ , ]

Applying Hermite-Hadamard Inequality,

( ) + ( )≥ −

( ) ≥+

⇒ ∫ ≥ ≥ + since, ≥ +

⇔ ≥ √ + [ . ≥ . ] ⇔( ) √

≥ (proved)

Solution 2 by Rozeta Atanasova-Skopje

< < ⇒ ≠ , − ≠ 0

− > ,∀ ∈ ℝ ∖ { } ⇒ > + 1 > √ + ⇒√ +

> 1

=−

( − ) √ +=

−( − )√ +

=− −−( − ) ⋅

√ +

> 1 ⋅ 1 = 1 =

208. Find , ∈ ℤ such that:

+ + = +

Proposed by Rovsen Pirguliyev-Sumgait-Azerbaidian

Solution by Santos Martins Junior-Brussels-Belgium

( , ) = ( , ) is the trivial solution

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( + + ) = ( + )( + ) and ( + ) = ( + ) −

Equation becomes: ( + )( + ) = ⋅ ( + ) −

Squarring: ( + ) ( + ) = (( + ) − )

= ( + ) { − ( + ) }

= ( − )( + ) (1)

Let = (2) where is integer ≠ since RHS is integer ⇒ − = (3)

(1) because: = ( − )( + ) ⇒ − +

⇒ = − + ⇒ = − +

Let = ⇒ = ( − )( + ) ⇒ + − + =

where roots are complex or irrational numbers ⇒ = is complex or irrational.

But from (3): ⋅ = + meaning that LHS is not interger ⇒ no integer

solutions.

209. Solve for real numbers:

⎩⎨

⎧ [ ] + ( − [ ]) =√

( − [ ]) + [ ] =, [∗]−



Squaring and adding: [ ] + ( − [ ]) + [ ] ( − [ ]) +

+ ( − [ ]) + [ ] + [ ] ( − [ ]) = +

+ ([ ] + − [ ]) = ; = → =

Solution 2 by Togrul Ehmedov-Baku-Azerbaidian

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⎩⎪⎨

⎪⎧ [ ] + { } + [ ] + { } =

√ +

[ ] − { } + [ ] − { } =√ −

⎩⎪⎨

⎪⎧ +

[ ] − { }=√ +

+[ ] − { }

=√ −

+ [ ] − { }

+ [ ] − { } =√ +√ −

→[ ] − { }

= −√ −√ +

[ ] { } = − → [ ] − { } = − ; ≤ [ ] + < 1 → − ≤ [ ] < 1 −

[ ] = − , ∈ ∅[ ] = , [ ] = → =


[ ] + ∙ ∙ ∙ … ∙ =

[ ] + ∙ ∙ ∙ … ∙ =,

[∗] - great integer function



= [ ] + , ≤ < 1; = [ ] + , ≤ < 1

∙ ∙ ∙ … ∙ =

∙ ∙ ∙ … ∙ =,

∙ ∙ ∙… ∙ ∈ ℤ, ≤ < 1 → ∙ ∙ ∙ … ∙ =

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∙ ∙ ∙… ∙ ∈ ℤ, ≤ < 1 → ∙ ∙ ∙ … ∙ =

: , ∈ , , , … , ( − )

211. Solve for > 0 the equation:

+ + + = ( ) + ( )



We know, for > 0, < ⇒ ( ) < ; ( > 1)

⇒ ( ) < ∴ ( ) > ∀ > 0, > 1

Thus, ( )

+( )

> + > + + +

Hence, given equation has no solution

Solution 2 by Soumava Chakraborty-Kolkata-India

( ) = ⇒ = > 1, ( ) = ⇒ = > 1

(∵ > 0)

Now, < , < and ∵ , > 1,∴ 0 < , < ⇒ < 2 , <

Now, given equation becomes: + + + = +

⇒ + + + = +

⇒+

++

= +

⇒ ( + ) + = +

⇒ ( ) ( ) = + ⇒ ⋅ ( ) ( )( )( )

= + ⇒ ( )( )( )

= + ⇒

+ = + (1)

∀ ∈ , , < ∴ as, < 2 , <

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< 2 ⇒ >( )

= and < 2 ⇒ >( )

=

(i) + (ii) ⇒ + > +

∴ LHS of (1) > of (1) ⇒ LHS never = RHS

⇒ no solution exists satisfying the given equation


√ − √ = √


Solution 1 by Abdallah Almalih-Damascus-Syria

Let = √ then the equation become: ( − ) =

so, − = , − = ; ( − ) =

if = then ± so, = √ = is root of this equation.

Let ≠ , divided by , − =

Let = hence ( − ) = , − + =

( − )( + − ) = ,( − ) ( + ) = so, − = , = , ⇒ =

the only root for this equation is = or + = , = − , + =

and the only root for this equation is =

we can take ( ) = ( ) + ; ( ) = + > 0

and →∓ ( ) = ∓∞, decreasing so, ( ) = ⇔ = unique

hence = = √ the only root of √ − √ = √

Solution 2 by Rajeev Rastogi-India

Given equation can be written as √ + √ + √ = ⋅ √ ⋅ √ ⋅ √

which is of the form + + = ⇒ either = = or + + =

⇒ either √ = √ or √ = − √ which has only one solution = .


Let √ = ( > 0) ∴ equation becomes:

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( − ) = ⇒ − + = (where = )

⇒ ( − ) ( + ) = ; = ⇒ =

Let ( ) = − ,∀ ≥ ∴ ( ) = − ≥

∴ ( ) ≥ ( ) ≥ ⇒ ≥ , equality at = ⇒ =

= − ⇒ = −

Let ( ) = + ,∀ ≥ ∴ ( ) = + ≥ > 0

∴ ∀ > 0, ( ) ≥ ( ) = ⇒ ≥ − , equality at =

∴ = − ⇒ = ⇒ = ∴ = is the only solution.


√ − √ = √



Solve the equation √ − √ = √

Let = √ then the equation become: ( − ) =

so, − = ; − = ; ( − ) =

if = then ± so, = √ = is root of this equation.

Let ≠ , divided by − =

Let = hence ( − ) = ; − + = ; ( − )( + − ) =

( − ) ( + ) = so, − = ; = ⇒ =

the only root for this equation is = or + = ; = −

+ = and the only root for this equation is =

we can take ( ) = ( ) + ; ( ) = + > 0 and →∓ ( ) = ∓∞

in decreasing so, ( ) = ⇔ = unique hence = = √ the only root of

√ − √ = √

Solution 2 by Rajeev Rastogi-India

Given equation can be written as √ + √ + √ = ⋅ √ ⋅ √ ⋅ √

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which is of the form + + = ⇒ either = = or + + =

⇒ either √ = √ or √ = − √ which has only one solution = .


Let √ = ( > 0) ∴ equation becomes:

( − ) = ⇒ − + = (where = )

⇒ ( − ) ( + ) = ; = ⇒ =

Let ( ) = − ,∀ ≥ ∴ ( ) = − ≥

∴ ( ) ≥ ( ) ≥ ⇒ ≥ , equality at = ∴ = ⇒ = ⇒

⇒ = ; = − ⇒ = −

Let ( ) = + ,∀ ≥ ∴ ( ) = + ≥ > 0

∴ ∀ > 0, ( ) ≥ ( ) = ⇒ ≥ − , equality at =

∴ = − ⇒ = ⇒ = ∴ = is the only solution.

Solution 4 by Athina Kalampoka-Greece

⋅ √ − √ = √ ; √ √ − √ = √

√ ⋅ √ − √ = √ ; ⋅ √ ⋅ √ ⋅ √ = √ + √

√ ⋅ √ ⋅ √ = √ + √ + √ . From here…

√ + √ + √ = or √ = √ = √

Therefore … √ = √ or √ = − √ . Therefore only one solution =

214. Solve the equation in ℝ:

√ + − + ( − )√ + = | + |.

Proposed by Nhuyen Van Nho-Nghe An-Vietnam

Solution 1 by Santos Martins Junior-Brussels-Belgium

We know that | | = for any real

Equation becomes: ( + − ) + ( − ) ⋅ ( + ) = {( + ) }

( + − ) − {( + ) } = −( − ) ⋅ ( + ) (1)

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Also we must have that ( + − ) ≥ ; we also have ( + ) ≥

Let ( + − ) = (2) and ( + ) = (3) where , ≥

Then ( − ) = ( + − ) − ( + ) ⇒ ( − ) = − (4)

Equation (1) becomes: − √ = − − ⋅ ( + )

− √ = − − √ + √ ( + + + + ) ⋅ ( + )

1) = √ is one solution ⇒ from (4): ( − ) = ⇒ =

Otherwise = − + √ ( + + + + ) ⋅ ( + )

⇒ IMPOSSIBLE since ≥ , ≥ , ( + ) > 0 meaning that RHS is ≤


Firstly, √ + − = √ + − ≥

Let us assume ≥ . Then, ( − )√ + ≥ ⇒ ≥ √ + − and,

∵ =

∴ | + | ≥ √ + − ⇒ ( + ) ≥ + − ⇒ ≥ ⇒ ≤ ⇒ only

possibility is = .

Now, let us assume ≤ . Then, ( − )√ + ≤ ⇒ ≤ √ + − and,

∵ = ,

∴ | + | ≤ √ + − ⇒ + + ≤ + − ⇒ ≥ ⇒ only

possibility is = . ∴ only solution is: = .


−[ ]

+ −[ ]

+ −[ ]

= [ ] + [ ] , [∗] - great integer

function



| − 1|[ ] + −[ ]

+ −[ ]

= [ ] + [ ] (1)

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Note that > 0, and RHS is not defined if [ ] = , + and RHS is negative

for [ ] = + , where ∈ ℤ. Also, RHS is equal to = if [ ] = + , ∈ ℤ

For = , [ ] = and (1) becomes,

− + − + − = (2)

Let − = (2) becomes

( ) = | + | + | | + | − | = ⇒ =

∴ = √ [∵ = ]

216.

Solve the equation in ℝ: √ + − + ( − )√ + = | + |.


Solution 1 by Santos Martins Junior-Brussels-Belgium

We know that | | = for any real

Equation becomes: ( + − ) + ( − )√ + = {( + ) }

( + − ) − {( + ) } = −( − )√ + (1)

Also we must have that ( + − ) ≥ ; we also have ( + ) ≥

Let ( + − ) = (2) and ( + ) = (3) where , ≥

Then ( − ) = ( + − ) − ( + ) ⇒ ( − ) = − (4)

Equation (1) becomes: − √ = − − ( + )

− √ = − − √ + √ ( + + + + ) ( + )

1) = √ is one solution ⇒ from (4): ( − ) = ⇒ =

Otherwise = − + √ ( + + + + ) ( + )

⇒ IMPOSSIBLE since ≥ , ≥ , ( + ) > 0 meaning that RHS is ≤


Firstly, √ + − = √ + − ≥ . Let us assume ≥ . Then ( −

+ ≥ ⇒ ≥ + − and, ∵ =

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∴ | + | ≥ + −

⇒ ( + ) ≥ + − ⇒ ≥ ⇒ ≤ ⇒ only possibility is = .

Now, let us assume ≤ . Then ( − )√ + ≤

⇒ ≤ √ + − and, ∵ = ,

∴ | + | ≤ + − ⇒ + + ≤ + − ⇒ ≥

⇒ only possibility is = ∴ only solution is: = .

217.

( ) = , ( ) = , ( ) = ( ) − ( ), ∈ ℕ, ≥

Find the reminder of , ≥ at − − +



Let ( ) = − − +

Suppose ( ) be the quotient and ( ) be the reminder when ( ) is divided by ( ), i.e., ( ) = ( ) ( ) + ( ). We assert

( ) = ( − ) + when = = ( − )( − ) +

when = − . For = , = ( ) − ,

( ) = ( ) = = ( )( − )( − ) +

( ) = ( ) = − = ( − ) + = ( − ) +

Thus, the result holds for all ∈ ℕ, ≤ , ≥ .

We prove the result for + . If + is odd, let + = + , where ≥ and

= , − = − . Now, as the result holds for , −

( ) = ( ) = ( )− ( ) =

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

= [ ( )− ( )] ( ) + ( )− ( )

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

= ( − ) + − ( − )( − )

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= [ ( ) + − ] + − ( − )( − ) −

= ( ) + ( − + )( − ) +

Thus, ( ) = ( ) = ( + )( − ) +

Next, when + is even, we let + = , − , − < , where ≥ .

Now, ( ) = ( ) = ( )− ( ) =

= [ ( ) ( ) + ( )]− [ ( ) ( ) + ( )]

= [ ( )− ( )] ( ) + ( )− ( )

But ( ) − ( )

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

= ( − )( − ) + − ( − ) ( − )−

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

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

∴ ( ) = [ − + − ( − + )]( − ) +

= ( − ) + = ( − ) + = ( + ) ( − ) +

Thus, the result holds in this case also. By the principle of mathematical induction

result is true for all ∈ ℕ.

218. Find all functions : ( ,∞) → ℝ such that:

+ √ + =


Solution by Abdallah El Farissi-Bechar-Algerie

Let = = √ , =

∙ ∙ + ∙ ∙ + ∙ ∙ =

( ) + ( ) + =

( ) = → ≡

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219. If : [ ,∞) → ℝ, ( ) + = ,∀ ≥ then ∀ , , ≥ :

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


Solution by Khanh Hung Vu-Vietnam

If : [ ; +∞) → ℝ, ( ) + = (1) ∀ ≥ then ∀ , , ≥ :

( ) + + ( ) + + ( ) + ≥ ⋅ ( − )( − )( − )

Substitute to , we have (1) ⇒ + = (2)

Substitute to , we have (1) ⇒ + ( ) = (3)

(2)+(3) ⇒ + + + ( ) = + ⇒

⇒ + = + ⇒ =( )

(4)

Substitute to , we have (4) ⇒ ( ) =( )

⇒ ( ) + = ∀ ≥

We have ( ) + + ( ) + + ( ) + = + +

By AM-GM inequality, we have + + ≥( )( )( )

So, ( ) + + ( ) + + ( ) + ≥( )( )( )

(QED)

The equality occurs when = = .

220. Find all functions :ℝ → ℝ such that:

+ ≤ ( ) + ( ) ≤ ( + ),∀ , ∈ ℝ


Solution by Chris Kyriazis-Greece

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Let’s set ( ) = ( ),∀ ∈ ℝ. Then the given inequality

+ ≤ ( ) + ( ) ≤ ( + ). For = = we have that:

≤ ( ) ≤ ( ) ⇒ ( ) = . For = − , we have that:

≤ ( ) + (− ) ≤ ( ) ⇒ ≤ ( ) + (− ) ≤ ( ),∀ ∈ ℝ ⇒

(− ) = − ( ),∀ ∈ ℝ (1)

For = : ≤ ( ) ≤ ( ),∀ ∈ ℝ (2)

We set → − and ...− ≤ (− ) ⇒( )− ≤ − ( ) ⇒ ≥ ( ),∀ ∈ ℝ (3)

So, (by (2)+(3)) ( ) = ,∀ ∈ ℝ. Then ( ) = ,∀ ∈ ℝ

which is acceptable because it verifies the given conditions.

221. Find :ℝ → ℝ such that: ( − ) ( ) − ( ) = ( + ) ( − )



When = = , we get = ( ) ⇒ ( ) = when = , we get

( ) = ( ) . If ( ) ≡ , we are done.

If ( ) ≠ for some ≠ , then = ( ). Thus, ( ) = ∀ or ( ) = ∀

Solution 2 by Bedri Hajrizi-Mitrovica-Kosovo

For = , we get = ( ) ⇒ ( ) =

For = : ( ) = ( )

( ) − ( ) = ⇒ ( ) = ,∀ ∨ ( ) =

For = : − ( ) = − (− )

Solution 3 by Khanh Hung Vu-Vietnam

Find :ℝ → ℝ such that: ( − ) [ ( )− ( )] = ( + ) ( − ) (1)

→ : (1) ⇒ = ( ) ⇒ ( ) =

→ : (1) ⇒ ( ) = ( ) ⇒ ( )[ − ( )] = (*)

Suppose that ∃ , ≠ such that ( ) = and ( ) =

→ , → : (1) ⇒ ( − ) − = ( + ) ( − ) (2)

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Case 1: ( − ) = : (2) ⇒ ( − ) = ⇒ = ⇒ = (Absurd)

Case 2: ( − ) = − : (2) ⇒ ( − ) − = ( + )( − ) ⇒

− = − ⇒ = (Absurd). So, (*) ⇒ ( ) = ∀ ≠ or ( ) = ∀ ≠

On the other hand, we have ( ) = . Then ( ) = ∀ ∈ ℝ or ( ) = ∀ ∈ ℝ

222. Choose a point inside circumcircle of , = , = , = .

Find the probability that this point its not in yellow area in terms of , , .


Solution by Khanh Hung Vu-Ho Chi Minh-Vietnam

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Construct the squares , and as shown.

Put , , is the center of ( ), ( ) and ( )

Put , , , is the radii of ( ), ( ) and ( ), ( )

1) Prove that ( ), ( ), ( ) are internally tangent circles with ( )

Since ∥ so ∃ : → , → ⇒ : →

On the other hand, and is the center of and ⇒ , , . Similarly,

we have , , and , , ⇒ ( ), ( ), ( ) are internally tangent circles with ( )

2) Prove that = , = and =

Put ⊥ . Put = . By Thales theorem for ∥ ⇒ =

Similarly, we have = ⇒ + = ⇒ + = ⇒ = =

Similarly, we have = and =

2) Prove that ( ), ( ), ( ) are 3 externally tangent circles

Since : → and and is the center of and ⇒

⇒ = ⇒ = = + ⇒ = + ⇒ = +

Similarly, we have = and = .

We need to prove that = + . By Cos’s law for , we have

∠ =+ −

⋅

Similarly, we have ∠ =⋅

. So, ⋅

=⋅

⇒ = + − ⋅ ⋅ ⋅+ −

⋅

⇒ ( + ) = ( − ) + ( − ) −( − )( − )

( + − )

⇒ + + = − ( + ) + + −( − )( − )( − )

⇒ = − ( + ) −( − )( − )( − )

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⇒ ⋅ + ⋅ + = − + + + −− + − + ( − )

⇒ ⋅ ( + )( + ) = − + + + −⋅ + ⋅ + ( − )

⇒ ( + )( + ) = −( + + )

( + )( + ) −( − )

( + )( + )

We have LHS = − =

⇒ ( + )( + ) =−

( + )( + ) ⇒

⇒ = − ⇒ = ⇒ = ⇒ = (True)

So, = +

⇒ ( ), ( ) are externally tangent circles. Similarly, we have ( ), ( ) and ( ), ( ) are

externally tangent circles. So ( ), ( ), ( ) are 3 externally tangent circles.

3) Put : “This point its not in yellow area in terms of , , ”. Find ( )

We have ( )

= ( ) ( ) ( )

( )=

⋅

⇒( )

= + + + + +

⇒ ( ) = − + + + + +

So ( ) = − + + vs =

223. Find:

=+ + − ( + + )+ + − ( + + )

Proposed by Daniel Sitaru-Romania

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Solution by Yen Tung Chung-Taichung-Taiwan

=+ + − ( + + )+ + − ( + + ) =

=− ( + )( + )( + )( + + + + + )

− ( + )( + )( + ) =

= ( + + + + + ) =

= + + + + + =

= + + + + + = + + =

224. Prove that:

( ) = , ( ) > 0

Proposed by Nawar Alasadi-Babylon-Iraq

Solution by proposer

∵→

− = ∴ =→

−

Let = ⇒ =

− = ( − ) ( ) = ( − )

=( − )

+( − )

( + ) ( − )

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=( − ) … ⋅ ⋅

( + )( + ) … ( + − ) =!

( + )( + ) … ( + )

∴→

− =→

!( + )( + ) … ( + )

∵ ( ) = ( − )!

∵ ( − )! =!

= ⋅⋅ ⋅ ⋅ … ⋅ ( + )( + ) … ( + )

( + )( + ) … ( + )

=! ( + )( + ) … ( + )( + )( + ) … ( + ) =

! ⋅ ( + ) ⋅ ( + ) … ( + )

( + )( + ) … ( + ) , →∞

∴→

! ( + ) ⋅ ( + ) … ( + )

( + )( + ) … ( + ) =→

!( + )( + ) … ( + )

∴ ( ) =→

!( + )( + ) … ( + )

∴ =→

− =→

!( + )( + ) … ( + ) = ( )

∴ ( ) = , ( ) > 0

225. Find:

=( + − ( + ) )( + − ( + ) )

(( + ) − − )



( + ) − − = + = ( + )

( + ) − − = + + +

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

= ( + )( + + )

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( + ) − − = + + + + +

= + + ( + ) + ( + )

= ( + )[ − + − + + ( + − ) + ]

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

=[− ( + )] − ( + ) + +

( + ) ( + + ) = =

Solution 2 by Yen Tung Chung-Taichung-Taiwan

Since + − ( + ) = − ( + )

+ − ( + ) = − ( + )( + + )

( + ) − − = ( + )( + + )

so, ∫ ∫ ( ) ( )

( )=

=− ( + ) (− ( + )( + + ) )

( + )( + + )

= = × ( − ) × ( − ) =

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

+ − ( + ) =

= ( + ) ( − + − + ) − ( + + + + )

= ( + )(− − − ) = − ( + )( + + )

+ − ( + ) =

= ( + ) − + − + − − ( + )

= ( + ) − − − − −

= − ( + )( + + + + ) = − ( + )( + + )

226. Find: = ∫ ⋅⋅

, ∈ ,


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= =( − )

( − )( − )

=( − )

( − )( − )

=− ( − )

( − )( − )

Put = ; − =

= −( − )

( − )( − ) = −−

( − )( − )

= − − − − = − − −

= | − | − | − | + = − − − +


=( − )

( − )( − )

=−

( − )( − ) ⋅

⇒

=

=−

( − )( − ) ⋅ −

=−

( − )( − ) = − − − =

= | − | − | − | +

= − − − + = | |− +

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227. Compute the following integral:

( ( − ) − )+ + ( + ) + +

Proposed by Mihaly Bencze-Romania

Solution 1 by Nitin Gurbani-India

∫ ( − )−

+ + + + +;

( − )−

+ + + + +

∫ ; let + + =

+= + = + + +


=( − )−

+ + ( + ) + +

Dividing numerator and denominator by

=− −

+ + + + +=

− −

+ + + + +

= ∫ . Put + + = ; − − =

=+

= + = + + +

228. Find:

=

⎝

⎜⎛

−

⎠

⎟⎞


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− =( − ) + ( − ) + ⋯+ ( − )

( − )( − ) … ( − )

=− ( + )

( − )( − ) … ( − ) = ( )− ( + )−

Where =( )( )…( )

( )( )( )…( )

( )

=( )!( ) ( )!

= ( )( )!

−− . Thus,

⎝

⎜⎛

−

⎠

⎟⎞

=− ( + )( − )!

(− ) −−

+ −+ −

229. If < ≤ ≤ then:

( + ) ∙ ( + ) ≥ ∙ ( + ) ∙ ( + )


Solution by Daniel Sitaru-Romania

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

ʹ=

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

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

230. If < < and > 0, then compute:

− + ( + )( − ) +( + ) ( ) + ( + )( + ) +

Proposed by Mihaly Bencze-Romania

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

(easy if you consider ( ) + ( + ) + )

So: ( ) ( )( )( ) ( ) ( )( )

=

=( ) − + ( + ) − + −

+ + + +=

=+ + − + + + + + − + +

+ + + +=

=+ + − + +

+ + + ++

+ +−

+ +=

=+ +

−+ +

++ +

−+ +

=

=−

+ +−

−

+ +=

+ +

+ +−

+ +

+ +

So ∫ ( ) = + , ∈ ℝ

231. Find:

… …

where { } represents the fractional part of

Proposed by Kays Tomy-Nador-Tunisia

Solution by Togrul Ehmedov-Baku-Azerbaidian

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… … = … …

= … { }

∑

∑

… = … …

= ∗ ! =!

232. If : [ , ] → ℝ, ( ) = , ( ) = is integrable and invertible then:

= [ ( ) + ( )] =

Proposed by Nishant Kumar-Jamsedpur-India


( ) = ( )| − ( ) ( ) = ( ) == ( ) =

= ( ) − ( ) − ( ) ( ) = × − − ( ) ( )

= − ( ) ( )

= ( ) ⇒ = ( ), = ( ) , = ⇒ == ⇒ =

( ) = ( ) ⋅ ⋅ ( ) = ( ) ( )

[ ( ) + ( )] = − ( ) ( ) + ( ) ( )

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= − ( ) ( ) + ( ) ( ) =

Solution 2 by Marian Ursarescu-Romania

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

for second integral = ⇒ = | ⇒ =

= ⇒ = ; = ⇒ =

= ( ) + ⋅ ( ) ⋅ = ( ) + ⋅ ( )

for second integral: ( ) = ⇒ = ( )

= ⇒ = ; = ⇒ = ; = ( )

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

= ( ) + ( ) | − ( ) = ( ) − ( ) = ⋅ − =

Solution 3 by Abdallah El Farissi-Bechar-Algerie

Let = ( ) ⇒ = ( )

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

= ( ) + ( ) ( ) = ( )

= ( )| = ( )− ( ) =

233.

For , ∈ ℕ∗ ∧ , ≥ . Prove:

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( ) ( )=

( ) ( )


Solution 1 by Abdallah El Farissi-Bechar Algerie

( ) ( )=

( ) ( )=

( ) ( )

=( )

⋅( )

+( ) ( )

=( ) ( )

Solution 2 by Marian Ursarescu-Romania

( ) ⋅ ( ) = ( ) ⋅ ( )

Let = ∫ ( ) ⋅ ( ) =

= ( ) ⋅ ( ) = (( ) ) ( )

= ( ) ( ) − ( ) ⋅ ( ) ⋅ (− )

= ( ) ⋅ ( ) ⋅ = ( ) ( )

234. Calculate:

=( ) + − + + + + ( + )( + − − )−

( − + )( − + )



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Numerator

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

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

−( − + ) − ( − + )

= + − ( − + ) + + − ( − + )

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

= ( + + + − )( − + )( − + )

Thus,

= ( + + + − ) = + + + −

= + + + − = + − ]

= + − =+ −

= −

235. Calculate:

=( ( ( ( ))) + )

( )( ( ))( ( ( )))


Solution by Igor Soposki-Skopje

=( + )

( )( )( ) ==

= ⋅ ⋅ ⋅

= ( + ) = + == = =

= =( + )

=( + )

=( ) − ( )

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236. Let be a positive integer. Evaluate:

( − ) − ( ) +( − )

Proposed by George Apostolopoulos-Messolonghi-Greece


Let = ∫ ( ) ( )

( ). For = , = . Let > 1.

( − ) − ( ) + = ( − ) ( − ) ( ) − ( ) − ( ) −⋯− − −

Thus

=( − ) ( ) − ∑

=( ) −

( )

=

= ∑ ∫( )

(1)

We now use, if is continuous for ≥ and → ( ) = , then for , > 0,

∫ ( ) ( ) = ( ( ) − ) . Let ( ) = , then ( ) = , = ,

∴ ∫ = (2)

Thus, from (1), (2)

=−

= ( !)

237. Find:

= −

Proposed by Arafat Rahman Akib-Bangladesh

Solution by Shivam Sharma-New Delhi-India

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⇒( )

− ⇒ ( ) ( )

⇒ ( ) ⇒( + )

( + ) ⇒( + )

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

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

⇒ ( ) ( ) + ( ) ( ) + ( ) ( ) + ( ) ( )

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

⇒ ( ) + − [ + ( ) − ( )] + + + + −

⇒ ( ) + + ( ) − ( ) − − ( ) + ( ) +

+ − + (OR)

= ( ) + ( ) [ − ] + − − ( ) [ − ]

238. Find:

= ( ( + ))

Proposed by Shafiqur Rahman-Bangladesh

Solution by Togrul Ehmedov-Baku-Azerbaidian

( ( + )) = ( ( ))

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=

⎣⎢⎢⎡

( ( ))

⎦⎥⎥⎤− + − ( ( ))

= ( ( )) − [ ( ( ))]

= ( ( )) = ( ( )) = −

= [ ( ( ))] = − [ ( ( ))]

= ( ( )) − [ ( ( ))] = [ ( ( ))]

=(− ) −

( ) =(− ) −

( )

=(− ) −

⎣⎢⎢⎡ ( )

−( )

⎦⎥⎥⎤

=(− ) −

⎣⎢⎢⎡− ( )

⎦⎥⎥⎤

=(− ) − ( )−

=(− ) −

=− (− )

=− (− )

= +(− )

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

239. Let ≥ be positive integer then prove that:

= ( + ) +( + ) −

+

Where {. } denotes the fractional part function and – zeta function

Proposed by Kays Tomy-Nador-Tunisia

Soltion by Togrul Ehmedov-Baku-Azerbaidian

= ={ }

={ }

+ { } =

={ }

+ =( − )

+ ( + )

=( − )

= ( + ) = ( + ) =

= ( ) = =

= − = − =

= − ! ( + )! =!

( + )! −

=!

( + )! =!

( + )!( )

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=!

( + )!!

( + ) =! !

( + )! ( + ) =! !

( + )![ ( + ) − ]

=( + ) −

+

= + ( + ) =( + )−

+ + ( + )

Note 1. ∫ = ! ∑( )!

Note 2. ∫ ( ) = ( )( )

= !( )

240. If , , > 0, = 1,

( ) =+ − + +− + − +

then:

( ) + ( ) + ( ) ≥ + ( + )


Solution 1 by Seyran Ibrahimov-Maasilli-Azerbaidian

( ) =+ − + +− + − + =

=( − + − + ) + ( − + )

− + − + =

= +( − + )

( + )( − + ) = +( − + )

( + )( − + ) =

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

( ) + ( ) + ( ) ≥ + ( + ) ( + )( + )

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( + ( + )) ≥ + ( + )( + )( + )

∑ ≥ √ ≥ and ∑ ( + ) = ∏( + )

( ) = + ( + ) ≥⏞

≥ √ + ( + ) = + ( + )


= ( − ) − = − − −

= − + − + ⋯ = ( − + − + ⋯ ) = +

= ( − ) − = − − = − −

= − + − + ⋯ = ( − + − + ⋯ ) = +

∴ − = + − + =( − )

( + )( + ) > 0 ⇒ >

241. If , , > 0, = 1

( , ) = ( + )( + )

then:

( + ) ( , ) + ( + ) ( , ) + ( + ) ( , ) > 6


Solution 1 by Dimitris Kastriotis-Athens-Greece

Let ( ) = ( + )( + ), ∈ [ , ], , > 0

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( ) =+ +

( + )( + )> 0 → , ( ) ≥ ( ) = √

( + )( + ) ≥ √ → ( , ) = ( + )( + ) > √

( + ) ( , ) > ( + )√ ≥⏞ √ ∙ √ =

= ≥⏞ ∙ ( ) =


Let ( ) = √ + , ( ) = √ + , ∈ [ , ]

≤ ≤ , ≤ + ≤ + → √ ≤ √ + ≤ √ +

√ ≤ √ + ≤ √ +

√ + ≥ √ , √ + ≥ √

( , ) = ( + )( + ) ≥⏞ √ + ∙ √ + ≥ √

( + ) ( , ) > ( + )√ ≥⏞ √ ∙ √ =

= ≥⏞ ∙ ( ) =

Solution 3 by Lazaros Zachariadis-Thesalonikki-Greece

( , ) = ( + )( + ) =

= √ + √ ( √ + √ ) ≥⏞ + √ =

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=12 + √ > √

( + ) ( , ) > ( + )√ ≥⏞ √ ∙ √ =

= ≥⏞ ∙ ( ) =

242. If , , ≥ ,

( , ) = ( ( + ) − )( ( + ) − )

then:

( ( , ) + ( , ) + ( , )) <


Solution by Dimitris Kastrotis-Athens-Greece

Let ( ) = ( + ) − , ( ) = − , ( ) = −( )

< 0, ≥ 1

− , ( ) ≤ ( ) = − < 0 → −

∈[ , )( ) = ( ) = → ( + ) − ≤ → ( + ) ≤

Equality holds for =

[ ( + )− ] ∙ [ ( + ) − ] =+

∙+

=

= + + ≤ ∙ ∙ ∙ =

( , ) = ( ( + ) − )( ( + ) − ) ≤

≤ ( ) = ∙

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( ( , ) + ( , ) + ( , )) < ∙ ( + + ) ≤

≤ ( + + ) =

243. If < , , ≤ ,

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

then:

( , ) + ( , ) + ( , ) ≥ √



+ + ≥ + , ≥+ + ≥ + , ≥

⎩⎪⎪⎨

⎪⎪⎧ + + ≥ + = +

+ + ≥ + = +

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

= + + + + ≥ +

( , ) ≥ + ≥

( , ) ≥ ≥⏞ ∙ ( ) ≥ √

Solution 2 by Lazaros Zachariades-Thessaloniki-Greece

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( , ) = ( + + )( + + ) ≥⏞

≥ ( + + ) =+

+ ; ( , ) ≥+

+ =

= + + + + + ≥⏞ ( ) = √ ≥

Solution 3 by Myagmarsuren Yadamsuren-Darkhan-Mongolia

( , ) = ( + + )( + + ) ≥⏞

≥ ( + + ) = + + =+

+

( , ) ≥+

+ =

= + + + + + ≥⏞ ( ) = √ ≥

244. Prove that:

( ) ( ) + ( ) ( ) >

Proposed by Uche Eliezer Okeke-Anambra-Nigeria


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

= ( ) ( ) ≥⏞+

+=

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( ) ( ) + ( ) ( ) > =

245. If , , > 0,

( , ) =( + + )

( + )( + )

then:

( , )+

( , )+

( , )≥


Solution by Dimitris Kastriotis-Athens-Greece

+ + ≥ ( + ) → ( + + ) ≥ ( + ) , ( )

( + )( + ) ≤⏞( + )

, ( )

( ), ( ) →( + + )

( + )( + ) ≥( + )

( + )= ( + ) ≥

( , ) =( + + )

( + )( + ) ≥ =

( , )≥ →

( , )+

( , )+

( , )≥

246. Prove that:

√ − √ + √ >

Proposed by Uche Eliezer Okeke-Anambra-Nigeria

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= − + √ = ( − ) √ =

= − = >⏞

> =√

=√√

∙ =

247. Prove that:

= + + + + + <


Solution by Dimitris Kastriotis-Athens-Greece

= ∙ ∙ =

= + + + + + = ∙ =

< ↔ < ↔ < 0.833333. . ( )

248. If , , ≥

( , ) = | ( − ) ( + ) − ( + )|

then:

( , ) + ( , ) + ( , ) ≤ √ ( + + )


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| ( − ) ( + ) − ( + )| ≤

≤ | ( − )| ∙ | ( + )| + | ( + )| ≤ | ( + )| + | ( + )| ≤

≤ √ ∙ ( + ) + ( + ) = √

( , ) ≤ √ = √

( , ) + ( , ) + ( , ) ≤ √ ( + + ) ≤ √ ( + + )

249. Prove that:

− <



= √ =√

⋅ = ( ) = − = − =+

≥ + ∀ ∈ ℝ ⇒ ≥ + ⇒ ≥ ( + )

⇒ ∫ ≥ + = ⇒ ∫ ≥ ⇒ − ∫ ≤ − : (1)

≤ √ ∀ ∈ [ , ] ⇒ ≤ √ ∀ ∈ [ , ] ⇒↑[ , ]

≤ √ ∀ ∈ [ , ]

⇒↑[ , ]

∫ ≤ ∫ √ ⇒ ∫ ≤ = : (2)

( )( ) ⇒ ∫ − ∫ ≤ − : (3)

− − = < 0 : (4)

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( )( ) ⇒ − <


For < < 1, 2 < 2 ⇒ ∫ < ∫ =

Also, for < < 1, > 0,

⇒ > 1 ⇒ > 1

∴ − <−

− =−

<

−< ⇔ − < 2 ⇔ < 9

250. If < ≤ then :

+ ++ +

≥( − )

Proposed by Nho Nguyen Van-Vietnam


+ ++ + =

( + + )( + − )+ + =

= ( + − ) =( − )( − )

−( − )

≥( − )

↔

↔( − )( − )

≥( − )

↔ ( + + ) ≥ ( + + ) ↔

↔ ( − ) ≥

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251. If ≥ then:

+ + + ≥


Solution by Myagmarsuren Yadamsuren-Darkhan-Mongolia

≥ then:

( + ) + ( + ) = ( + )( + ) ≥

≥ √ + = + √ ⋅ √ =

= √ ⋅ + √ ⋅ = ⋅ ⋅ + ⋅ √ =

= ⋅ ⋅ ⋅ +⋅

= ⋅ ⋅ ⋅ + ⋅ = ⋅

252. If < ≤ then:

( + )+

+( + )

+≤ ( + )



≤ − → ≤ − → ( + ) ≤−

( + ) →

( + ) ≤ ( + ) →

( + ) − ( + ) ≤ ( + ) →

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( ( + ) ( + )) ≤ ( + ) →

( + )+ +

( + )+ ≤ ( + )

253. If ≤ ≤ ≤ then:

√ ( − ) ≤ + ≤√ ( − )

Proposed by Nho Nguyen Van-Vietnam


By Schweitzer’s inequality:

( + ) + ≤( + )∙ ∙ ∙ → + + ≤ → + ≤ =

√

+ ≥ √ → √ ≤ + ≤√

→

√ ≤ + ≤√

√ ( − ) ≤ + ≤√ ( − )

254. If < < < 1 then:

( − )( + )( + )

( ) ( ) > 1


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For < , < 1 ( + )( + )

( ) ( ) =( + )( + )

= ( + )+

≥ = [ ]

= [ ] . But ≥ ∀ > 0

[ ( ) = , > 0, ( ) = ( + ), ( ) < 0 if < <

> 0 if > , = if = ]. Thus,

=( + )( + )

( ) ( ) ≥ ⇒ ≥

Now, > 1 ⇔ ( ) > 1 ∴ > 1

⇒ > ( − ) = ( − ) ⇒ ( − ) > 1

255. If , , ≥ then:

+ + √√ − + − + √ −

≥


Solution by Myagmarsuren Yadamsuren-Darkhan-Mongolia

= ( − ) + ⋅ ( − ) + ≥ ⋅ √ − + − + √ −

⋅∑∑√ −

≥

≥ ⋅⋅ ∑√ −∑√ −

= ⋅ ( − )( − )( − ) =

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256.

= , = ,

=

Prove that: − + ≤



For ≤ , , ≤ ; ( − )( − )( − ) ≤

⇒ − + ≤

Also, < , ,

∴ − + ≤

⇒ −

+ ≤ ⇒ − + ≤ =

257. If < ≤ < 1 then:

√ √ +≤ −


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√ √ +≤ √ =

√≤

≤+ √

≤⏞√

= = ∙ − = −

258. If , , > 1 then:

( ) ⋅ ( ) ≤ + +



( ) ⋅ ( ) ≤

≤⋅ + ⋅

+ =

= ⋅ ≤ ⋅+

=

⋅ ( ) ⋅ ( ) ≤ ⋅ =

⋅ () ≤ =

=++ ≤

+= + + = + +

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( ) = ( ) ( )

≤( )( ) + ( )( )

+ =( + )

≤

≤ ( )√ = . Thus, ∫ ( ) ≤ ∫ = . But ≤ + as

≤ +

∴ ( ) ≤ + ⇒ ( ) ≤ + +

259. If , , ≥ then:

+≤ ( )( )


Solution by Soumitra Mandal-Chandar Nagore-India

+ ≤

= ≤ ∵ ≥

= = ( − )( − ) = ( )( )

260. If < ≤ then:

+ +

√ + +≥ −


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+ +

√ + +=√ −√ −

∙√ −

√ −=√ +√ +

= √ + − ≥

+ +

√ + +≥ = √ = −

261. If < ≤ then:

+ −≥ ( − )



≥ +≥ + → + ≥ + + ≥⏞ + →

+ − ≥ →+ −

> 2 →+ −

> →

+ −≥ = ( − )

262. For ≤ ≤ ≤ . Prove:

( − ) ≤+

++

≤√

( − )


Solution by Soumitra Mandal-Chandar Nagore-India

≤ ≤ ≤ ,+

++

=( + ) + ( − )

≥ +

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++

+≥ + ≥ = ( − )

++

+= + + + − ≤ + +

++

+≤ + +

= ( − ) + ( ) , need to prove, ( − ) + ( ) ≤ √ ( − )

⇔−− ≤

√ −≈ .

applyin MVT = ≤ since ≤ ≤ ≤ ≤

hence ≤ √ , which is true. Hence proved.

263. For < 0

Prove: ∫ + ≥


Solution 1 by Chris Kyriazis-Greece

Set ( ) = + , ≤ ≤ , < 0

is continuous on [ , ] and ( ) = ⋅ + ⋅ −

( ) = ⇔ = √ (since ≤ ≤ ), so

√

( ) + + + + + + + + + + + + + + + + + − − −− −− −− −−− −− −−

( )

local minimum at = : ( ) =


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maximum at = √ : √ = ⋅

Since is continuous at closed interval [ , ] the smallest local minimum is the

minimum of the function. So, ( ) ≥ ( ),∀ ∈ [ , ]

⇒ ( ) > ( ) ⇒ + ≥ ⋅


Let ≤ but ≠ and be a rational number which does not lie between and

then ( − ) ≥ −

∫ + = ∫ [let – = > 0]

≥ ∫ ( ) [since, ≤ ≤ ⇒ + ≤ + ≤ + ]

= we need to prove, ≥

⇔ − ≥ − , which is true. Hence proved.

264. Prove that:

( − )

−≤

−≤



≥⏞ = ∙∙ − ∙ = ( − )

≤ = −

265.

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If ≤ , , , ≤ then:

√ + √ + √+

≤


Solution 1 by Nho Nguyen Van-Nghe An-Vietnam

⊕ , , , ∈ [ ; ]

→− √ − ≥− √ − √ ≥

− − √ ≥→

√ + ≤ +

√ + √ ≤ + √+ √ ≤ +

→ √ + √ + √ ≤ + + √ = + + √ +

≤ + →√ + √ + √

+≤

So: ≤ ∫ ∫ ∫ ∫ = (done)


If ≤ ≤ ⇒ √ ≤ ⇒ − √ ≥ So, − ≥ , − √ ≥ , − √ ≥

We have that: − √ − ≥ ⇒ + ≥ √ + (1)

And − √ − √ ≥ ⇒ + √ ≥ √ + √ (2)

So, by (1) + (2) : √ + √ + √ ≤ + √ + √ (3)

So, it suffices to prove that

+ + √ ≤ + or + + √ + ≤ + or

+ ≥ + √ . But this one holds because:

≤ ⇒ − ≥√ ≤ ⇒ −√ ≥

⇒ − −√ ≥ ⇒ + ≥ + √

So, √ + √ + √ ≤ + ⇔ √ √ √ ≤

Integrating this one we have what we want!

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266. Prove that:

( + + ) ≤



( + + ) ≤ ( + + ) ≤

≤ ( + + ) =

= + + = + + =

267.

For < 0. Prove: ∫ + ≥



Set ( ) = + , ≤ ≤ , < 0

is continuous on [ , ] and ( ) = ⋅ + ⋅ −

( ) = ⇔ = √ (since ≤ ≤ ), so

√

( ) + + + + + + + + + + + + + + + + + − − −− −− −− −−− −− −−

( )



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maximum at = √ : √ = ⋅

Since is continuous at closed interval [ , ] the smallest local minimum is the

minimum of the function. So, ( ) ≥ ( ),∀ ∈ [ , ]

⇒ ( ) > ( ) ⇒ + ≥ ⋅


Let ≤ but ≠ and be a rational number which does not lie between and

then ( − ) ≥ −

∫ + = ∫ [let – = > 0]

≥ ∫ ( ) [since, ≤ ≤ ⇒ + ≤ + ≤ + ]

= we need to prove, ≥

⇔ − ≥ − , which is true. Hence proved.

Solution 3 by Manish Moryani-India

To prove: For > 0, ∫ + ≥ . Let ( ) = + , ( ) = −

∴ ( ) > 0∀ < −√ & > √ ∴ in [ , ], [ ( )]( ) = + =

Let = − ⇒ > 0[∵ < 0] ∴ ∫ + = ∫

+ ≤ ; ≥ ∴ > for > 0

=+

≥ ≥ ⋅ [ ] =

268.

For < ≤ . Prove ∫ ∫ ( + ) − ≥ ( − ) .

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Solution 1 by Lazaros Zachariadis-Thessaloniki-Greece

+ = ( + )( − + − + )

++ = − + − +

( + ) −++ = + + + + − + − + −

= + + = ( + + ) ≥ ⋅ ( + )

= ( + ) ≥ ⋅ =

≥ = = ( − )

= ( − ) = ( − )

Solution 2 by Sanong Huayrerai-Nakon Pathom-Thailand

( + ) −++ =

( + ) − +( + )

=( + + + )

( + )

=( + + + )

( + ) ≥( )( + )

≥ ( ) = ] = ( − )( − ) = ( − )


( + ) = + ( + ) + +

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( + ) −++ =

( + ) ++

= [ ( − + ) + ] = ( + + )

= + +

= ( − ) ( + ) + ( − )

we need to prove, ( − ) ( + ) + ( − ) ≥ ( − )

⇔ ( − ) ( + ) ≥ ( − ) ⇔ ( − ) ( + + ) ≥

Hence proved.

269.

For < ≤ . Prove ∫ ∫ ( )( ) + ( )( ) ≥ √ ( )( )

.


Solution by Serban George Florin-Romania

+( + )( + ) +

+( + )( + ) ≥

+( + )( + ) ⋅

+( + )( + ) =

=( + )( + )

≥( + )( + )

=√

( + ),

+( + )( + ) +

+( + )( + ) ≥

√( + )

=

=√ ( − )

( + ), ≥ .

270. Prove:

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√ + + √ + √+ + + +

≤



We have that (By A.M-GM) ≥ √ ⋅ ⋅ ⋅ ⋅ ⇒ + ≥ √

Doing the same: + ≥+ ≥ √+ ≥ √

. Adding these inequalities, we have that,

+ + + + ≥ √ + + √ + √ ⇔√ + + √ + √

+ + + + ≤

Integrating this inequality we take we want!

271. Prove:

+ + ++ + +

≥



Let ( ) = , ≤ ≤ ; + + = ≥

≥( + ) + ( + ) + ( + )

≥

≥( ) + ( ) + ( )

= ( ) + ( ) + ( ) ≥( + + )

so, ∫ ∫ ∫ ≥ ∫ ∫ ∫( )

= ( + + ) ≥ =

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easy to prove strictly decreasing = ( ) =

( ) =( − ) ⋅ ( + )

( + )

′ −− −− −− −− −− − + + + + + +

( ) = = , ( ) ≥ ∀ ∈ [ , ]

, , ∈ [ , ], ≤ ≤ , ≤ ≤ , ≤ ≤

(+) −− −−

≤ + + ≤


+ + ≥ + + | ⋅

( + ) ≥ ⋅ ≥ ⋅ ⇔ + ≥ ⋅ |(+ )

∑ + ≥ ⋅ ∑ + ⇔ ∑∑

≥ (*)

∑ +∑ + ≥ = ⋅ ⋅ =

272. If < ≤ , : [ , ] → ℝ, ( ) > 0,∀ ∈ [ , ] then:

( ) ≥ ( )



Function is strictly increasing in [ , ] so does ( ) = , ∈ [ , ].

It suffices to prove that

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( − )∫ ( ) ≥ ( − ) ∫ ( )

or ( ) ∫ ( ) ≥ ( − )∫ ( )

or − ∫ ( ) ≥ ( − ) ∫ ( )

or ∫ ⋅ ∫ ( ) ≥ ( − )∫ ( )

That is true due to Chebyshev’s inequality, because is strictly decreasing in

[ , ], > 0

(easy to check) and , ( ) are strictly increasing in [ , ].

273. For > 0. Prove:

+ + +

+ + +≤ √



+ + +≤

+ + +⇒

( + + + )+ + + ≤

⇒+ + +

( + + + )≤ ( ) ⇒

+ + +

( + + + )

≤ ( ) = ( )

274. Find:

= → ∫[ ] [ ] ⋯ [ ]

, [∗] - great integer function

Proposed by Rovsen Pirguliyev-Baku-Azerbaidian


For ≤ ≤ , ∈ ℕ, let ( ) = [ ] + [ ] + ⋯+ [ ]

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For ≤ < , [ ] = [ ] = ⋯ = [ ] =

⇒ ( ) = + + ⋯+ for ≤ < ; > + + ⋯+

= >

for ≤ < . We now show that = ∫ ( ) = ∞ for each given .

For a given ∈ ℕ, write = + where = ∫ ( ) , = ∫ ( )

Note > 0, therefore > ≥ ∫ = ∫ = ∞

∴ → ∞ ⇒ → = ∞

275. Find:

=→

− −


Solution by Henry Ricardo-New York-USA

We have

= + + + = + + +

Thus

= + + +

= +( + )

+( + )( + )

++ +

= + + ++ +

+

and

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− − = + + ,

so that

− − = + + →

as → ∞.

276. For > 0. Prove:

+ ++

+ ++

+ +≥


Solution by Lazaros Zachariadis-Thessaloniki-Greece

≥ = so,

+ + ≥ ⋅ + + + + +

≥ ⋅ ⋅( )

∏( + ) = ⋅∏( + )

< ≤ ≤ 2< ≤ ≤ 2< ≤ ≤ 2

⇒ ≤ ≤

⇒ ≤ + ≤ ⇒ ≤ ( + ) ≤ ⋅

⇒ ≤ ( + ) ≤ ⇒∏( + )

≥

≥

-----------------------------

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∏( + )≥ = ⇒

∏( + )≥

thus ∑

≥ = ⋅ ⋅ ⋅ ⋅ =

277.

Prove: ∫ + + √ ≥ + √



∀ ∈ [ , ] it’s √ ≥ "=” for = , = ⇔ √ + ≥

⇔ √ + ≥ √ ⋅ √ ≥ √ ⋅ ⇔ √ + + ≥ √ + ⋅

⇔ √ + + ≥ √ + ⋅ √ so,

+ √ + ≥ √ + ⋅ √ = √ + ⋅ ⋅ = √ + ⋅

278. Let > > 1 and be a positive integer. Prove that:

√+ ( ) + ⋯+ + +

≤ ( ∈ ℝ).

Proposed by George Apostolopoulos-Messolonghi-Greece


+ ( ) + ⋯+ + + ≥ ( + ) ⋅ ( ) … ⋅ ⋅

= ( + ) ( ) ⋯ = ( + ) ⋅ ( ⋯ )

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= ( + ) ⋅⋅ ⋅( )

= ( + ) ⋅⋅( )

= ( + ) ⋅ so,

√ ⋅+ ⋯+ + + ≤

( + ) ⋅

= + = + ⋅ [ − ] = + ⋅ =

279. Prove that:

+ > +



I will use the well-known inequality ≥ + , ∀ ∈ ℝ

= ≥ ⋅ + ; = ≥ +

Adding those two inequalities, we have that

+ ≥ ( + ) + ⇒

( + ) > ( + ) + ⇒

( + ) > ⋅ [ − ] + ⇒ ( + ) > 2 +

280. Prove that:

( + + )( + + ) ≤


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( + + )( + + ) ≥⏞ ∙ =

( + + )( + + ) ≤ → ( + + )( + + ) ≤

( + + )( + + ) ≤ =

281. If : [ , ] → ℝ, continuous then:

+ ( ) > ( )



Using Cauchy – Schwarz inequality for integrals, we have that:

( ) ≥ ( ) ⇒ ( ) ≥ ( ) ⇒

+ ∫ ( ) ≥ + ∫ ( ) .So, it suffices to prove that:

+ ∫ ( ) > ∫ ( ) (1)

Setting ∫ ( ) = , its easy to see that (1) holds, because:

+ > ⇔ − + > 0 for every ∈ ℝ ⇔ ( ) − + > 0 or

− + > 0 for every ∈ ℝ. ; = − < 0 so, this is true!

282. If > 0, ∈ ℝ then:

+ ≤

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The function ( ) = is convex on ( ,∞) then

+ =+

≤ ( + )

we have ≤ ( + ) ≤ , ( , ∈ [ , ]) and ≤

then

= ≤ + ≤ =

It follow that ≤ ∫ ∫ ≤

Solution 2 by Rozeta Atanasova-Skopje

+ ≤

=

= ⋅ − ⋅ − = ( − )( − )

=− ( + ) +

=+

≤


+

+≤

++ ⇒ + ≤ +

⇒+

≤ +

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=+

= < 1

283. If : [ , ] → ( ,∞) continuous then:

+ ( ) ( ) ≤ ( )


Solution 1 by Chris Kyriazis-Athens-Greece

Using AM-GM we have that

+ ( ) ( ) ≥ ( ) ( ) (1) So, by GM-HM we have

( ) ( ) ≥( ) ( )

⇒ ( ) ( ) ≥( ) ( )

(2)

So, using (1)+(2) we have that

+ ( ) ( ) ≥( ) + ( )

⇔( )

( )

+ ( ) ( ) ≤( ) + ( )

⇒

+ ( ) ( ) ≤( ) + ( )

⇒

+ ( ) ( ) ≤∫ ( ) + ∫ ( )

=∫ ( )


+ ( ) ( ) ≤ ( ) ( )=

( )≤ ( )


+ ( ) ( ) ≤( ) ( )

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≤ ( ) ⋅ ( ) = ( ) ( )

= ( )

284. If , > 1 then:

++ ≤

⋅



≤ ⋅ ≤ ⋅ = so,

++ ≤ + = + −

= ⋅ [ + − ] = ( + − − )

= + − ( ) = ⋅ − ( )

= ⋅⋅

=


++ ≤ + ≤

≤ ⋅ ⋅ = ( − ) ⋅ ( − )

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≤( − ) + ( − )

=⋅ − ( )

=

285.

If ≤ < then:

+ + + + >


Solution 1 by Khanh Hung Vu-Vietnam

Put ( ) = + + + when ≤ <

We have ≤ ⇒ ≤ ⇒ ≤

Lemma: + ≥ ( ) when > 0, > 0

Applying the lemma, we have

( ) ≥+ + +

⇒ ( ) ≥+

Since ≤ ⇒ ≥ ⇒ + ≥

So, ( ) ≥ ( ) ⇒ ( ) ≥ ⇒

⇒ + + + ≥

⇒ + + + ≥ −

⇒ + + + + ≥ ⇒

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⇒ + + + + >

(QED). The equality doesn’t occur.


+ + + + ≥

≥ + ⋅ + + + =

≥ + + + ≥ +

+( + )

+ = + ⋅ =

= + ⋅ − = + − =


+ + + +

≥ + + ∵ ≥ ≥ = + −

= + − ≥ + − = >

286. If : [ , ] → [ ,∞), – continuous, then:

( ) + ( ) ≥ ( )


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Solution 1 by Abdallah El Farisi-Bechar-Algerie

– continuous by Cauchy – Schwarz

( ) ≤ ( )

( ) + ( ) = ( ) ( ) + ≥ ( ) ≥ ( )

Solution 2 by Khanh Hung Vu-Ho Chi Minh-Vietnam

If : [ ; ] → [ ; +∞), – continuous, prove that

∫ ( ) + ∫ ( ) ≥ ∫ ( ) (1)

By BCS inequality for integral, we have

∫ ( ) ⋅ ∫ ( ) ≥ ∫ ( ) (2)

On the other hand, by the BCS inequality, we have

∫ ( ) ⋅ ∫ ≥ ∫ ( ) ⇒ ∫ ( ) ≥ ∫ ( ) (3)

(2) and (3) ⇒ ∫ ( ) ⋅ ∫ ( ) ≥ ∫ ( ) ⇒ ∫ ( ) ≥ ∫ ( ) ⇒

⇒ ∫ ( ) + ∫ ( ) ≥ ∫ ( ) + ∫ ( ) (4)

By AM-GM inequality, we have

( ) + ( ) ≥ ( ) ⋅ ( ) ⇒

⇒ ∫ ( ) + ∫ ( ) ≥ ∫ ( ) (5)

(4) and (5) ⇒ (1) true ⇒ QED

The equality occurs when ( ) and ( ) are proportional,

∫ ( ) = ∫ ( ) , for example ( ) =

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Using the generalized Hölder’s inequality for three functions we have that:

( ( ) ≥ )

( ) ≤ ( )

⇒ ∫ ( ) ≥ ∫ ( ) (1)

So, it suffices to prove that

( ) − ( ) + ( ) ≥

But ∫ ( ) − ∫ ( ) + ∫ ( ) ≥( )

( ) − ( ) + ( ) ≥

( ) ( ) − ( ) + ≥ ⇒

( ) ⋅ ( ) − ≥

which holds! ( ( ) ≥ )


: [ , ] → [ , +∞): [ , ] → ∫ ( ) = ≥ ∶ (1)

By the Hölder Inequality we deduce

[ ( )] ⋅ ⋅ ≤ ( )

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⇒ ∫ ( ) ≥ ∫ ( ) ⇒ ∫ ( ) ≥ ∫ ( ) = ∶ (2)

( ) + ( ) ≥ ( )

⇒ ( ) + ( ) − ( ) ≥

= ( ) + ( ) − ( )

≥{( ),( )}

+ − ≥ √ ⋅ − = − = (true)

287. If < ≤ ≤ + 3, : [ , ] → ( ,∞), ( ) > 0,∀ ∈ [ , ] then:

( ) ≤ ( ) + +

Proposed by Nishant Kumar-Jamshedpur-India

Solution by proposer

Using Cauchy-Schwarz inequality we get:

(∫ ( ) )2 ≤ ∫ ∫ ( ) = ∫ ( )

It is given that function in increasing in the interval [ a , b ] ,

which implies f(x) ≤ f(b) for all x ≤ b. Squaring we get f2(x) ≤ f2(b) for all x ≤ b

We can say that ∫ ( ) ≤ ∫ ( ) = ( − ) ( )

(∫ ( ) )2 ≤ ∫ ∫ ( ) =

= −

( − ) ( ) ≤ ( + + ) ( )

288. If < ≤ < then:

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+ + + ≥ ( + ) − ( + )


Solution by Khanh Hung Vu-Ho Chi Minh-Vietnam

Lemma: + + + ≥⋅ √

∀ ∈ ;

Surely, we have 1) > 0 ⇒⋅ √

< 2) > 0, ≤ and ≤ ⇒

+ + + > + + + > . So,

+ + + ≥⋅ √

. Applying the lemma, we have

+ + + ≥√ +

⇒ + + + ≥ ( + ) − ( + )

(Q.E.D). The equality occurs when = .

289. Prove that:

√ + + √ + √ +

√ + + √ + √≥



√ + = √ + √ + + + ≥⏞ √ ∙ √ ∙ ∙ ∙ = √

( √ + ) ≥ √ → √ + ≥ √ →

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→√ + + √ + √ +

√ + + √ + √≥ →

√ + + √ + √ +√ + + √ + √

≥ =

290. Prove:

++

++

+<


Solution 1 by Dimitris Kastrotis-Athens-Greece

= {( , , ) ∈ : ≤ , , ≤ }; =

+ ≥ → + ≤ → + ≤ → + ≤

→ + ≤ ≤

→ + ≤

→ + ≤ ⋅ = < =


+ + + + + ≤+

++

++

=+ +

so, ∑ ≤ ( ) ≤ ( + + ) thus,

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( + + ) = ⋅ + +

= + + = ⋅ + +

= + + = ⋅ + = ⋅ =

+ ≤


For , > 0, ≤

∴ + + + + + ≤+

++

++

= ( + + ) = [( + + ) − ( + + ) ]

= [( + ) − ( + ) − ( + ) + ( + ) ]

= − − + − + + − = <

Solution 4 by Serban George Florin-Romania

≤ , + ≤+

, + ≤+

, + ≤( + )

+ ≤ + ≤ ( + ) =

= ⋅ ( + + + + + ) =( + + + + + )

≤

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≤( + + + + + )

=⋅ ( + + )

=⋅ ( + + )

,

+ + ≥ + + ⇒

⇒ + ≤ ( + + )

( + + ) = + ( + ) =

= + + = + + =

= + + = + + = + + = =

⇒ + ≤ ⋅ =

Solution 5 by Sanong Huayrerai-Nakon Pathom-Thailand

+ + + + + ≤+

++

++

≤+ +

=+ +

=( + + )

=( + + ) − ( + + )

= =( × × − )

=

Therefore is to be true.

291. If , , , :ℝ → ( ,∞) continuous functions, ≥ then:

( ) ≥ ( ) ( ) ( ) ( ) − ( ) ( ) ( )


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Solution by Chris Kyriazis-Athens-Greece

We have that:

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

( ) ( ) ( ) ( ) = ( ) ( ) ( ) ( )

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

Integrate from to , we have what we want. ( > )

(if = , it’s obvious)

292. If , , ∈ ℕ∗then:

( − ) ≥( !)

( + + )!



( , ) = ( − )

( − ) ≥ ( − ) [∴ ≥ ,∀ ≥ ] =

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

( + + ) ∴ ( , ) =( ) ( )( + ) ≥⏞

≥( ( + ))( + + ) =

( !)( + + )!


If = = = → = ∫ − − ≤ ∫ < 1 = ( !)( )!

=

As ≥ ,∀ ∈ [ , ] → ∫ ( − ) ≥ ∫ ( − ) =

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

( + + )

( − ) ≥

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

( + + ) ≥⏞

≥( ( + ))( + + ) =

( !)( + + )!

293. In , , , – angles. Find:

( ) =→

( + ) ( + ) ( + )( + ) ( + ) ( + )

, ∈ ℝ



Let ( , ) = ( + ) ( + ) ( + )( + ) ( + ) ( + )

Using → − ( ) ; → + ( ) , we get

( , ) = =

[∴ and are proportional]. Now,

( , ) = ⇒ ( ) =→

( , ) =

294. Find:

=→

+ + + ⋯+


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=→

+ + + ⋯+=

=→

=→

= =

Solution 2 by Shivam Sharma-New Delhi-India

=→

+ + + ⋯+=

=→

= =

295. Find:

=→ ( + )

∙ ( + − ) +



= ( + − ) + = ( + − ) + − + =

= ( + − ) + − − + =

= ( + − ) + − − + =

= ( + ) − ( + ) =

=→ ( + ) ∙ =

→ ( + ) ∙ =

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296. Find:

=→

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



Let = ( )…( )( )( )…( )

= !!( )!

=( )( )

( )!!( )!

= ( + )( + )+ = ( + )( + ) ⋅

⋅ − + − + − + − + − +

= ( + )( + ) − − ( + )−( + )( + )

= ( + )( + ) − ( + )− ( + )( + )

= ( + )( + )−( + )

( + )( + ) −

=→ ( + )( + ) −

+

+ ( + )−

=→ ( + )( + ) − ⋅

+

+ ( + )− =

= − − =

Solution 2 by Henry Ricardo-New York-USA

First we note that ( ) = ( + ) = ∑ . Now we find that

( ) =( + )

+ − + = − + .

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Integrating with respect to from to gives us

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

=⋅ ( − ) ⋅ ( − ) … ( − + )

⋅ ⋅ ⋅… ⋅

Dividing by , we see that

=→

⋅ ( − )( − ) … ( − + )⋅ ⋅ ⋅… ⋅ =

→ ( + )( + ) + ( ) = .

297. If , , > 0 find:

=→


Solution 1 by Rajsekhar Azaad-Gaya-India

⇒→

( ( − ))( ( − ))

⇒→

( ( − ))( ( − )) ×

⇒ ( )× × × ⋅ ×( )

( )× × × × ×( )×

(L. Hospital)

⇒→

( − )( − ) ×

( − )( − ) = × =


Let = → , → ; → = ; = ; = ⋅

= ⋅ ⋅ = ⋅ ⋅⋅

; = ⋅

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= → , → ; → =

= , =⋅

= ⋅ ⋅ = ⋅ ⋅⋅

; = ⋅

= ( ) = = ; ( ) = ; =

= ( ) = = ; ( ) = ; =

=→

⋅ = = ⋅→

⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅

= ⋅→

⋅ = ⋅→

⋅⋅ = ⋅

⋅⋅ = ⋅

⋅⋅ =


=→

=→

( ) ( )( )

( ) ( )( )=

( ) ⋅

( ) ⋅=

298. Find:

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=→

+( + )( + )( + ) +



For > 1, let ( ) = ∫ ( )( )( )( )

= ∫ ( )

Put + = .

( ) = ( + ) + = ( + ) + − =√ −

+√ −

=√ −

+ +√ −

−√ −

→( ) = ∵ > 0, 0 < ( ) <

299. Find:

=→

√ ! ⋅ √ ! ⋅ … ⋅ ( )!( )!



We know for > 1; ( !) < ⋯ =

∴ ( !) ( !) ⋅ … ⋅ ( !) <+ +

⋅ … ⋅+

= ⋅( + )!

⇒( !) ⋅ ( !) ⋅ … ⋅ ( )!

( )! <( + )!

( )! ⋅ ⇒ <( !) ( !) ⋅… ⋅ ( )!

( )! <

As → = , we get →( !) ( !) ⋅…⋅ ( )!

( )!=

300. Find:

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=→

∑ ∑

√ !



We have ∑ = − ; ∑ − = − −

and we know from Stirling formula ! ∼ √ ⋅ ; √ ! ∼ √ ⋅

so →∑ ∑

√ != →

√ ⋅= ∞


For ≥

= − = ( − ) −

= − ( + ) > + − ( + )

= ( + ) − = ( − )( + )

Also, ( !) < ⋯ =

∴∑ ∑

( !)>

( − )( + )+ > − 1,∀ ≥ 2

⇒→

∑ ∑

( !)= ∞

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Its nice to be important but more important its to be nice.

At this paper works a TEAM.

This is RMM TEAM.

To be continued!

Daniel Sitaru

rmm - calculus marathon 201 - 300 - ssmrmh.ro · calculus marathon 201 – 300 . ... khanh hung...

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