matsefi/octogon/volumes/octogon_2017_2_proposed... · proposed problems 475 proposed problems...

Proposed Problems 475

Proposed problems

PP27064. 37 Solve the following system:

x+ y + z + t = (x+ y − z)2

x2 + y2 + z2 + t2 = (t+ z − x+ 1)2

x3 + y3 + z3 + t3 = (4x+ 5)3.

Mihaly Bencze

PP27065. Solve the following system:

�2x2 + 2y2 + z2 = (y + z)2

2 · x4 + 2y4 + z4 = (x+ z)4

Mihaly Bencze

PP27066. Solve in Z the following system:

�x2 + y2 + z2 = (z + 5)2

x3 + y3 + z3 = (z + 1)3.

Mihaly Bencze


x+ y + z =�

z70

�2x2 + y2 + z2 = (5x+ 535)2

x3 + y3 + z3 = (z +√x+ y + z)3

.

Mihaly Bencze

PP27068. Show that:�a (a+ 3b)2 − b (3a+ b)2

�·

·�a�a3 + 21a2b+ 35ab2 + 7b3

�2 − b�7a3 + 35a2b+ 21ab2 + b3

�2�=

=�a�a2 + 10ab+ 5b2

�2 − b�5a2 + 10ab+ b2

�2�2.

Mihaly Bencze

PP27069. Solve in N the equation�x41 + y41 + t41 + u41

� �x42 + y42 + t42 + u42

�= 74998.

Mihaly Bencze

37Solution should be mailed to editor until 30.12.2020. No problem is ever permanently

closed. The editor is always pleased to consider for publication new solutions or new in sights

on past problems.

476 Octogon Mathematical Magazine, Vol. 25, No.2, October 2017

PP27070. Solve in N the equationx�1050x + 1400y + 1430z + 1665t + 1562u

�=

= 735a + 3220b + 3780c + 4160d + 5936e.

Mihaly Bencze

PP27071. Determine all a, b, c, d, e, f, g ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for which

�aab

�a+d+ (aac)a+d +

�ade

�a+d+�ada

�a+d+�add

�a+d=

�dfge

�d.

Mihaly Bencze

PP27072. Solve in N the equation(1x + 2x + 3x + ...+ 24x)y = (15y + 16y + 17y + ...+ 34y)x .

Mihaly Bencze

PP27073. Determine all a, b, c, d, e ∈ {0, 1, 2, ..., 9} for which�ab�b

+ (ac)b +�ad

�b= (aec)a .

Mihaly Bencze

PP27074. Prove that8�a4 + b4 + (a+ b)4

��a8 + b8 +

�a2 + b2

�4��a16 + b16 +

�a4 + b4

�4�=

=�a2 + b2 + (a+ b)2

�2 �a4 + b4 +

�a2 + b2

�2�2·�a8 + b8 +

�a4 + b4

�2�2.

Mihaly Bencze

PP27075. Determine all a, b, c, d, e, f, g, h, k ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} forwhich (aa)c +

�ab�c

+ (ac)c +�ad

�c=

�be�c

and

(ca)c + (cc)c +�cf

�c+ (cg)c +

�ch

�c+�da

�c=

�kk

�c.

Mihaly Bencze

PP27076. If x, y, z > 0, thenP x5

(5x+3y)y4+ 1

27 · (�

x2z)5

9(xyz)5+5(xyz)4(�

x2z)≥ 1

8

P (xz+y2)5

(yz)4(6yz+5(xz+y2)).

Mihaly Bencze


PP27077. Denote Fk and Lk the kth Fibonacci, respective Lucas number.

Prove that min

(P

cyclic

F 21

3F2F3+5 5√

F 21 F

42 F

43

;P

cyclic

L21

3L2L3+5 5√

L21L

42L

43

)≥ n

8 .

Mihaly Bencze

PP27078. Solve in Z the equation�x3 + 3y3 + 4z3 + t3

�2017=

�2u4 + 4v4 + 13w4

�2017

Mihaly Bencze

PP27079. If a, b, c > 0 thenP a5

3+5a + (a+b+c)5

27(9+5(a+b+c)) ≥ 18

P (a+b)5

6+5(a+b) .

Mihaly Bencze


�x1 + x2 + x3 + x4 + x5 = 5x20171 + x20172 + x20173 + x20174 + x20175 = 5

.

Mihaly Bencze

PP27081. Solve in Z the equation x4 + y4 + z4 = 2 + t4 + u4.

Mihaly Bencze


�x2 + y2 + z2 = t2 + u2 + v2

x6 + y6 + z6 = t6 + u6 + v6.

Mihaly Bencze

PP27083. Solve in Z the equation�x41 + y41 + z41 + t41 + u41

� �x42 + y42 + z42 + t42 + u42

�=

=�x4 + y4 + z4 + t4 + u4

�2.

Mihaly Bencze

PP27084. Solve in Z the equation x2 + y2 = 2 + z2.

Mihaly Bencze


PP27085. Solve the equation�x3 + (x+ 1)3 + (x+ 2)3 + (x+ 3)3

�·

·�(x+ 20)3 + (x+ 22)3 + (x+ 24)3 + (x+ 26)3 + (x+ 28)3 + (x+ 30)3

�=

=�x3 − x

�3.

Mihaly Bencze

PP27086. Solve the equation(2x− 1)3 + (2x+ 2)3 + (2x+ 4)3 + (2x+ 6)3 + (2x+ 8)3 = 7x3.

Mihaly Bencze

PP27087. Solve the equation�x3 + (x+ 1)3 + (x+ 2)3

�·

·�(x+ 95)3 + (x+ 96)3 + (x+ 97)3 + (x+ 98)3 + (x+ 99)3

�=

=�2x4 + 3x3 + 6x2 + 17x+ 12

�2.

Mihaly Bencze

PP27088. Solve the equation(x− 6)4 + (x− 4)4 + (x− 2)4 + x4 + (x+ 2)4 + (x+ 4)4 + ...+ (x+ 120)4 =

=�2x3 + 9x2 − 19x− 6

�2.

Mihaly Bencze

PP27089. Solve in Z the following equation(x+ y)

�x2 + y2

�+�x3 + y3

�2= z2 + t2.

Mihaly Bencze

PP27090. Solve the equation�(x− 3)3 + (x− 2)3 + (x− 1)3

�·

·�(2x+ 1)3 + 8x3 + (2x+ 1)3 + 8 (x+ 1)3

�·

·�(5x+ 1)3 + (5x+ 3)3 + (6x− 1)3 + (6x+ 1)3 + 27 (2x+ 1)3 + (7x+ 1)3

�=

=�x2

�x3 + 4

��3.

Mihaly Bencze

PP27091. Solve in Z the equation�2x2 + y2 + 2z2

� �2x4 + y4 + 2z4

�= (y + z)2 (x+ y)4 .

Mihaly Bencze


PP27092. Solve in Z the following system:�x2 + y2 + z2 + t2

� �x4 + y4 + z4 + t4

�=

= (2 (x+ y) + 2683)2�

x+ty+z + 31156

�4.

Mihaly Bencze

PP27093. Solve in Z the equation(x+ y + z)

�x3 + y3 + z3

�+�x2 + y2 + z2

�2= u2v3 + t2.

Mihaly Bencze

PP27094. Solve the equation�4 (x+ 1)2 + (4x− 1)2 + (11x+ 4)2

��8 (x+ 1)3 + (4x− 1)3 + (11x+ 4)3

�=

= 9 (4x+ 1)2 (12x− 1)3 .

Mihaly Bencze

PP27095. Solve the equation�(2x+ 3)5 + (9x+ 2)5 + 177858x5

�· ((x− 8)5 + (x− 7)5 + (x− 6)5+

+(x− 5)5 + (x− 4)5 + (x− 3)5 + (x− 1)5) = x15.

Mihaly Bencze

PP27096. Solve the equationx4 + (22x+ 40)4 + (40x+ 130)4 + (42x+ 50)4 == (16x+ 120)4 + (21x+ 100)4 + (33x+ 100)4 + (46x+ 70)4 .

Mihaly Bencze

PP27097. Solve the equation�x2 + (2x+ 29)2

��x2 + (5x+ 20)2

��(2x+ 29)2 + (5x+ 20)2

�=

= (2x+ 37)2 (5x+ 24)2 (6x+ 3)4 .

Mihaly Bencze

PP27098. Solve the equation�(x+ 2)2 +

�x2 − 1

�2��(8x+ 1)2 +

�(6x− 1)2 − 1

�2�·


·�(47x+ 4)2 +

�(34x+ 1)2 − 1

�2�

= x4 (6x− 1)4 (33x+ 4)4 .

Mihaly Bencze

PP27099. Solve the equation(1x + 3x) (1y + 3y + 5y + 7y) (1z + 3z + 5z + ...+ 15z) = 1128881664.

Mihaly Bencze

PP27100. Solve the equation25x12−5280x11+92160x10−10x9+10560x8+110x6−1056x5−20x3+1 = 0.

Mihaly Bencze

PP27101. Solve the equation�(x− 6)2 + (x− 3)2 + (x− 2)2

��16x2 + (5x+ 6)2 + (2x− 1)2

�·

�1024x2 + (9x+ 6)2 + (9x+ 13)2

�= 5x.

Mihaly Bencze

PP27102. Denote Pk the kth Pell number. ComputeqP1 +

3pP2 +

4√P3 + ....

Mihaly Bencze

PP27103. Denote Fk the kth Fibonacci number. Compute F1

1+F2

1+F3

1+...

.

Mihaly Bencze

PP27104. Denote Fk the kth Fibonacci number. ComputeqF1 +

3pF2 +

4√F3 + ....

Mihaly Bencze

PP27105. Denote Lk the kth Lucas number. ComputeqL1 +

3pL2 +

4√L3 + ....

Mihaly Bencze


PP27106. Solve in Z the equation�x2 + y2 + z2

� �x3 + y3 + z3

�= (z + 1)3 (z + 5)2 .

Mihaly Bencze

PP27107. Solve in Z the equation�x2 + y2 + z2 + t2

� �x3 + y3 + z3 + t3

�= (t+ 1)3 (t+ 5)2 .

Mihaly Bencze

PP27108. Solve in Z the equation113

�x3 + y3 + z3

�= a3 + b3 + c3 + d3 + e3 + f3.

Mihaly Bencze

PP27109. Solve in Z the equation70 (1x + 2x + 3x + ...+ 24x) = 15y + 16y + ...+ 34y.

Mihaly Bencze


x2 + y2 = (y + 8)2

x2 + z2 = (z + 4)2

y2 + z2 = (z + 27)2.

Mihaly Bencze

PP27111. Solve in Z the equation�(x− 1)3 + (x− 2)3 + (x− 3)3 + (x− 4)3

�·

·�(2x+ 1)3 + (2x+ 3)3 + (2x+ 5)3 + (2x+ 7)3 + (2x+ 9)3 + (2x+ 11)3

�=

=�5x2 + 13x

�3.

Mihaly Bencze

PP27112. Solve in N the following system:�x3 + y3 = ab

z31 + z32 + z33 + z34 + z35 + z36 + z37 + z38 = bxab

Mihaly Bencze

PP27113. Solve in Z the equation�4x2 + y4 + 9z2

� �x4 · y12 + (2yz + 1)2 y6 + (z + y)2 · z4

�=


=�y6x4 + x2y10 + (z + y)2 z2

�2.

Mihaly Bencze

PP27114. Prove thatnP

k=1

�nk

�kk (n− k + 1)n−k−1 = n (n+ 1)n−1 .

Mihaly Bencze

PP27115. Prove that∞Pk=1

1k2(k+1)(k+2)(k+3)

= 136

�π2 − 49

6

�.

Mihaly Bencze

PP27116. If z1, z2, z3, z4 ∈ C such that z31 + z32 + z33 + z44 = 0. Determine alln ∈ N for which

P(1 + |z1|n) zn1 = 0.

Mihaly Bencze


k=0

�4n+1k

��3n−k2n

�= 22n+1

�2n−1n

�.

Mihaly Bencze

PP27118. Compute limn→∞

n3

�12 −

nPk=1

3k2+14k6+3k4+3k2−1

�.

Mihaly Bencze

PP27119. Let be M =nn2 sin2 x + n2 cos2 x|n ∈ N, n ≥ 2, x ∈ R

o. Prove

that cardM = n2 − 2n+ 2.

Mihaly Bencze

PP27120. Solve in R the following system:

log2�1 + 3

√x1

�= log7 x2

log2�1 + 3

√x2

�= log7 x3

−−−−−−−−−−−log2

�1 + 3

√xn

�= log7 x1

.

Mihaly Bencze


PP27121. If ai > 0 (i = 1, 2, ...,m), thennP

k=1

1m�

i=1

1ai+k

≥ n+ nm�

i=1

1ai

.

Mihaly Bencze

PP27122. In all triangle ABC holds 12 + 8P a

b+c +�

a3

abc ≤ (P

a)2P 1

ab .

Mihaly Bencze

PP27123. In all triangle ABC holds

9r2

s

P ctg2 A2

s+3r(ctgA2+ctgB

2 )≥ 1 + 16

P (ctgA2−ctgB

2 )2

(ctgA2+4ctgB

2+4ctgC

2 )(4ctgA2+ctgB

2+4ctgC

2 ).

Mihaly Bencze

PP27124. If xk > 0 (k = 1, 2, ..., n) and n ∈ N , m ≥ 1,nQ

k=1

xk = 1, then

nQk=1

1+xm+1k

1+xmk

≥ 1.

Mihaly Bencze

PP27125. In all triangle ABC holdsP �

tgA2 tg

B2

� 92 + 6

�rs

�3 ≤P�tgA

2 tgB2

� 72 .

Mihaly Bencze

PP27126. If abc = 1 where a, b, c > 0 thenP �ab2 + ac+ c

� �a2b+ bc+ c

�≥ 27

Mihaly Bencze

PP27127. In all triangle ABC holdsrna

an−1 + (rb+rc)n

(b+c)n−1 ≥ 4R+r2s

�rn−1a

an−2 + (rb+rc)n−1

(b+c)n−2

�, for all n ≥ 2, n ∈ N.

Mihaly Bencze

PP27128. In all triangle ABC holdsP hna

(tgB2tgC

2 )n−1 ≥ s2+r2+4Rr

2R

P hn−1a

(tgB2tgC

2 )n−2 for all n ≥ 2, n ∈ N.

Mihaly Bencze


PP27129. If ak > 0 (k = 1, 2, ..., n) , thennP

k=1

ank ≥ 2−nQ

cyclic

(a1 + a2) + (n− 1)nQ

k=1

ak.

Mihaly Bencze

PP27130. Determine all x, y, z, t ∈ N for which 3x + 4y + 5z + 6t is aperfect square.

Mihaly Bencze


�x2 − 24y2 = 1y2 − 2017z2 = 1

.

Mihaly Bencze

PP27132. In all triangle ABC holdsP 1

2a2+2c2−b2≤ s2−r2−4Rr

6s2r2.

Mihaly Bencze

PP27133. Prove that 66∞Pk=1

L8k+4

(F8k+1)(F8k+8+1) + 720∞Pk=1

F8k+4

(L8k+1)(L8k+8+1) =

= 105∞Pk=1

L8k+5

(F8k+1+1)(F8k+9+1) + 100∞Pk=1

F4k+4

(L4k+2+2)(F4k+6+2) .

Mihaly Bencze

PP27134. Prove that

18

� ∞Pk=1

F6k+3

(F6k+1)(F6k+6+1) +∞Pk=1

L6k+3

(L6k−9)(L6k+6−9)

�= 35

∞Pk=1

F4k+2

L4kL4k+4.

Mihaly Bencze


270r4 ≥�1 + r3

P 1r3a

� �40Rr + 10r2 − 3s2

�.

Mihaly Bencze

PP27136. Denote Fn and Ln the nth Fibonacci respective Lucas numbers.

Prove that 4∞Pi=1

F6i+3

F6iF6i+6= 9

∞Pi=1

L6i+3

L6iL6i+6.

Mihaly Bencze


PP27137. If a1 = 1 and 3an = 4an−1 +p7a2n − 3 then for all p ≥ 3 prime

exist n ∈ N such that an ≤ p ≤ an+1.

Mihaly Bencze

PP27138. In all triangle ABC holds 135R2r2

s2(5r2+20Rr−s2)≥ 1 + r3

P 1h3a.

Mihaly Bencze

PP27139. Determine all n for which 4n and 5n have the first two decimalsequal.

Mihaly Bencze

PP27140. If x > 0 then 2√6+

√3−4

√2√

6+ 1√

2shx+ 2

chx ≥ 4√1+ch2x

.

Mihaly Bencze

PP27141. Solve the following equations:

1).nP

k=1

tg kπ2n+1 = 0 2).

nPk=1

tg k2π2n+1 = 0 3).

nPk=1

tg k3π2n+1 = 0

Mihaly Bencze

PP27142. Solve in Z the equation xyy+z2

+ yzz+x2 + zx

x+y2= 239

56 .

Mihaly Bencze


x+ y2 − z3 = 133x4 + y3 + z2 = 53x7 + y5 + z5 = −2883

.

Mihaly Bencze

PP27144. Determine all polynomials P ∈ R [x] for which

P (x)P�3xk − 2

�= P

�xk

�P (3x− 2) for all x ∈ R, when k ≥ 2, k ∈ N.

Mihaly Bencze


PP27145. If ak > 0 (k = 1, 2, ..., n) such thatnP

k=1

ak = 1 and {b1, b2, ..., bn}

is a rearranget of the set {a1, a2, ..., an} thennQ

k=1

aa2kk b

b2kk ≥

�nP

k=1

akbk

�2

.

Mihaly Bencze

PP27146. In all tetrahedron ABCD holds

1).Q�

rha

��

rha

�2

≥ r2P 1

h2a

2).Q�

r2ra

��

r2ra

�2

≥ r2

4

P 1r2a

Mihaly Bencze

PP27147. Prove thatnQ

k=1

�n+1

nk(k+1)

��

n+1nk(k+1)

�2

≥ 1n+1 .

Mihaly Bencze

PP27148. In all triangle ABC holdsQ�

2R4R+r cos

2 A2

�2( 2R4R+r

cos2 A2 )

2

≥ (4R+r)2−s2

2(4R+r)2.

Mihaly Bencze


rra

��

rra

�2

≥ s2−r2−4Rr2s2

.

Mihaly Bencze


rra

��

rra

�2

≥ 1− 2r(4R+r)s2

.

Mihaly Bencze


2R2R−r sin

2 A2

�2( 2R2R−r

sin2 A2 )

2

≥�8R2+r2−s2

2(2R−r)2

�.

Mihaly Bencze

PP27152. If M ∈ R [x] is a polynomial which have not real roots, then forany polynomials P ∈ R [x] exist polynomial Q ∈ R [x] and exist n ∈ N∗ suchthat (P (x))n + (Q (x))n is divisible by M (x) .

Mihaly Bencze


PP27153. In all triangle ABC holdsQ �

tgA2 tg

B2

�2(tgA2tgB

2 )2

≥�1− 2r(4R+r)

s2

�2.

Mihaly Bencze

PP27154. Prove that exist infinitely many triangles ABC for which√sinA,

√sinB,

√sinC ∈ Q.

Mihaly Bencze

PP27155. Determine all triangles ABC for which3√sinA, 3

√sinB, 3

√sinC ∈ Q.

Mihaly Bencze

PP27156. Prove that exist infinitely many triangles ABC for which√sinA,

√sinB,

√sinC ∈ R\Q.

Mihaly Bencze

PP27157. Determine all triangles ABC for which3√sinA, 3

√sinB, 3

√sinC ∈ R\Q.

Mihaly Bencze

PP27158. If p, q, r are prime thenmax{p,q,r}P

k=1

p

qk +

qpk + r

√k ∈ R\Q.

Mihaly Bencze

PP27159. Prove that exist infinitely many triangles A1A2A3 for whichsinA1

cosA2+cosA3; sinA2cosA3+cosA1

; sinA3cosA1+cosA2

∈ Q.

Mihaly Bencze

PP27160. Prove that exist infinitely many triangles A1A2A3 for whichsinA1

cosA2+cosA3; sinA2cosA3+cosA1

; sinA3cosA1+cosA2

∈ R\Q.

Mihaly Bencze

PP27161. Prove that exist infinitely many triangles A1A2A3 for whichcosA1

sinA2+sinA3; cosA2sinA3+sinA1

; cosA3sinA1+sinA2

∈ Q.

Mihaly Bencze


PP27162. Prove that exist infinitely many triangles A1A2A3 for whichcosA1

sinA2+sinA3; cosA2sinA3+sinA1

; cosA3sinA1+sinA2

∈ R\Q.

Mihaly Bencze

PP27163. 1). Prove that exist infinitely many triangles ABC for whichtgA, tgB, tgC ∈ Q2). Prove that exist infinitely many triangles ABC for whichtgA, tgB, tgC ∈ R\Q3). Prove that exist infinitely many triangles ABC for which tgA, tgB, tgCare transcendental numbers

Mihaly Bencze

PP27164. 1). Prove that exist infinitely many triangles ABC for whichsinA, cosB, tgC ∈ Q2). Prove that exist infinitely many triangles ABC for whichsinA, cosB, tgC ∈ R\Q3). Prove that exist infinitely many triangles ABC for whichsinA, cosB, tgC are transcendental numbers

Mihaly Bencze

PP27165. 1). Prove that exist infinitely many triangles ABC for whichsinA, sinB, sinC ∈ Q2). Prove that exist infinitely many triangles ABC for whichsinA, sinB, sinC ∈ R\Q3). Prove that exist infinitely many triangles ABC for whichsinA, sinB, sinC are transcendental numbers

Mihaly Bencze

PP27166. 1). Prove that exist infinitely many triangles ABC for whichcosA, cosB, cosC ∈ Q2). Prove that exist infinitely many triangles ABC for whichcosA, cosB, cosC ∈ R\Q3). Prove that exist infinitely many triangles ABC for whichcosA, cosB, cosC are transcendental numbers

Mihaly Bencze


PP27167. Calculate: I =R x− (x− 1)2 ln (x− 1)

x3 − 2x2 + xsin (lnx) dx for x > 0

Mihaela Berindeanu

PP27168. Calculate E = limn→∞

n

�1

2018 − n1R0

xn+1

x2+2017dx

�

Mihaela Berindeanu

PP27169. If a, b, c,m, n, p > 0; 2mn > p then:�m2a2

bc +n2bc−pa�3

+�m2b2

ca +n2ca−pb�3

+�m2c2

ab +n2ab−pc�3

≥ 3(2mn−p)3abc

D.M. Batinetu-Giurgiu, Neculai Stanciu

PP27170. If a, b, c,m, n > 0 then: (a2+2bc)2

mb+nc + (b2+2ca)2

mc+na + (c2+2ab)2

ma+nb ≥ (a+b+c)3

m+n


PP27171. If a, b, c, x, y ∈�0, π2

�then:

tan2 a+tan b tan ca2(b sinx+c sin y)

+ tan2 b+tan c tan ab2(c sinx+a sin y)

+ tan2 c+tan a tan bc2(a sinx+b sin y)

> 18(x+y)(a+b+c)


PP27172. If in ΔABC; s = 1;m ∈ N then:

3m+�a cot A

2

�m+1+�b cot B

2

�m+1+�c cot C

2

�m+1≥ 18(m+ 1)r


PP27173. If x, y, z ∈�0, π2

�;x+ y + z = π then:

1(sinx+sin y)2 sin z

+ 1(sin y+sin z)2 sinx

+ 1(sin z+sinx) sin y > 27

4π3


PP27174. If x, y > 0 then in ΔABC: a3

xb+yc +b3

xc+ya + c3

xa+yb ≥ 4√3S

x+y

D.M. Batinetu-Giurgiu, Daniel Sitaru

PP27175. If a, b, c,m, n > 0;x, y, z ∈ (0, 1) then:a

(mb+nc)x(1−x2)+ b

(mc+na)y(1−y2)+ c

(ma+nb)z(1−z2)≥ 9

√3

2(m+n)



PP27176. If k,m ∈ N∗; k < m then find a ∈ R such that the sequence(xn)n≥1; xn = 1

nk+1+ 1

nk+2+ . . .+ 1

nm − a log n is a convergent one.


PP27177. If a, b, c > 0;x, y ∈�0, π2

�then:

a2 + bc

a2(b sinx+ c sin y)+

b2 + ca

b2(c sinx+ a sin y)+

c2 + ab

c2(a sinx+ b sin y)≥

≥ 18

(x+ y)(a+ b+ c)


PP27178. If m ≥ 0;x, y, z > 0 then:�(xy)m+1+(yz)m+1+(zx)m+1

� 1

(x+y)2(m+1)+1

(y+z)2(m+1)+1

(z+x)2(m+1)

!≥ 9

4m+1


PP27179. Let I be the incentre of ΔABC and Ra, Rb, Rc the circumradii ofΔBIC;ΔAIC respectively ΔAIB. Prove that:

(Ra +Rb +Rc)�

RaRbRc

+ RbRcRa

+ RcRaRb

�≥ 12− 6r

R

Daniel Sitaru

PP27180. Prove that in any triangle ABC the following relationship holds:

16P�

mamc

+ mbmc

�4> 81

�ama

�4+�

bmb

�4+�

cmc

�4!

Daniel Sitaru

PP27181. Prove that if a, b, c, d, e > 0; a 6= b 6= c 6= d 6= e 6= a then:P a2

(b+c+d+e)(a−b)(a−c)(a−d)(a−e) <(a+b+c+d+e)2

1024abcde

Daniel Sitaru

PP27182. IfA = 2

√2 sin3 10◦ +

√2 sin 10◦ + 1

B = 2√2 cos3 20◦ +

√2 cos 20◦ + 1

C = 2√2 cos3 20◦ +

√2 cos 40◦ + 1


then ABC < 27

Daniel Sitaru

PP27183. Prove that if a, b, c ∈ R thenP |(a+ b)(1− ab)| < 32 +

Pa2 + 1

2

Pa4

Daniel Sitaru

PP27184. Prove that if a, b, c ∈ R thenP |(a+ b)(1− ab)| < 3

2 + a2+ b2+ c2

Daniel Sitaru

PP27185. Prove that if a, b, c, d ∈ R then:2(ad− bc)4 + 2(ac+ bd)4 ≥ (a2 + b2)2(c2 + d2)2

Daniel Sitaru

PP27186. Prove that in an ABC acute-angled triangle the followingrelationship holds:

PtanA tanB + 45 ≤ 2 tan2A tan2B + tan2C

Daniel Sitaru

PP27187. Prove that if x, y ∈ R; 0 < a ≤ b ≤ c then:(9a+ 12b+ 18c)(x2 + y2) + (18a+ 12b)xy ≥ (a+ b+ c)(13x2 + 10xy + 13y2)

Daniel Sitaru

PP27188. In ΔABC;O - circumcentre; I - incentre; G - centroid. Provethat:P �

S[AGB]S + S

S[AGB]

�2+�S[AIB]

S + SS[AIB]

�2+�S[AOB]

S + SS[AOB]

�2!≥ 100

Daniel Sitaru

PP27189. Let Ra, Rb, Rc be the circumradius of △BOC,△AOCrespectively △AOB,O - circumcentre of △ABC. Prove that:R2

aRb

+R2

bRc

+ R2c

Ra≥ 3R

Daniel Sitaru


PP27190. If x ≤ y ≤ −2 ≤ z ≤ t then:xex + yey + zez + tet ≥ (x+ y + 2)ex+y+2 + (z + t− 2)

3√ez+t−2

Daniel Sitaru

PP27191. In ΔABC: 9a2b2c2P

an−2bn−2 ≤ 4�P

an��P

anm2a);n ≥ 2

Daniel Sitaru

PP27192. Prove that if a, b, c > 0; abc = 1 then:�Pa4��P a

b

��Pa3��P a

c

��Pa2�≥�P

a�3�P 1

a

�2

Daniel Sitaru

PP27193. Let Ωa,Ωb,Ωc be the circumradii of ΔBGC,ΔCGA respectivelyΔAGB, G - the centroid of ΔABC. Prove that: 27ΩaΩbΩc ≥ 4Rs2

Daniel Sitaru

PP27194. Prove that in any triangle ABC: max (a,b,c)r ≥ 3

p24√3

Daniel Sitaru

PP27195. Prove that if: a, b, c > 0; a+ b+ c = 3 then:P ac1+a+ab +

P bc1+b+ab +

P ac1+a+a2

≤ 3

Daniel Sitaru

PP27196. Prove that if x, y, z > 0;x2 + y2 + z2 = 12; then:

P

xy+ y

x1x+ 1

y

!+P

11x+ 1

y

!≤ 9

Daniel Sitaru

PP27197. If x ∈ (0,π); y > 0 then: 2y+sin2 x

+ 2y2+sinx

≤ 1sinx

√sinx

+ 1y√y

Daniel Sitaru

PP27198. Prove that in any ABC triangle the following relationship holds:P (a2−ab+b2)2

a2+4ab+b2≥ 2S√

3

Daniel Sitaru


PP27199. Prove that in any ABC triangle the following relationship holds:

Q

m8a+m8

b

m6a+m6

b· l8a+l8bl6a+l6b

· h8a+h8

b

h6a+h6

b

!≥

2S2

R

!6

Daniel Sitaru

PP27200. Prove that if x, y, z > 0; 6xyz = 1x+2y+3z then:

(4x2y2+1)(36y2+z2+1)(9x2z2+1)2304x2y2z2

≥ 1(x+2y+3z)2

Daniel Sitaru

PP27201. Prove that in any triangle ABC the following relationship holds:P (a2−ab+b2)2

a2+4ab+b2≥ 2S√

3

Daniel Sitaru

PP27202. Prove that in any ABC triangle the following relationship holds:(a3+p)(b3+p)(c3+p)

(a2b+3√3r)(b2c+3

√3r)(c2a+3

√3r)

≥ 1

Daniel Sitaru

PP27203. If a, b, c, d > 0 then:(a+ c)c(b+ d)d(c+ d)c+d ≤ cc · dd · (a+ b+ c+ d)c+d

Daniel Sitaru

PP27204. Prove that in any ABC triangle:a11

b+c−a + b11

c+a−b +c11

a+b−c ≥ a10 + b10 + c10

Daniel Sitaru

PP27205. Prove that if a, b, c > 0 then:P�

ab

�2·P

�ab

�4·P

�ab

�8≥P

�ac

�·P

�ba

�·P

�bc

�

Daniel Sitaru

PP27206. Prove that if A,B ∈ Mn(C) such that AB = C anddet(C2 + C + In) 6= 0 then BABA+BA+ In is invertible.

Daniel Sitaru


PP27207. Prove that if A,B ∈ Mn(C) such that AB = C anddet(C3 +C2 +C + In) 6= 0 then BABABA+BABA+BA+ In is invertible.

Daniel Sitaru

PP27208. Prove that if A,B ∈ Mn(C) such that AB = C anddet(Cm + Cm−1 + ...+ C2 + C + In) 6= 0,m ∈ N;m ≥ 3 then:(BA)m + (BA)m−1 + (BA)m−2 + ...+BA+ In is invertible.

Daniel Sitaru

PP27209. Let be a, b ∈ (1,∞); fixed. Study the series’ convergence:xn = n

p(log2 a)

n + (log2 b)n

Daniel Sitaru

PP27210. Let be A =

2 ε ε2

ε 5 ii ε2 7

; ε = −1

2 + i√32 . Prove that it exists

B,C ∈ M3(C) such that: A = B2015 + C2016

Daniel Sitaru

PP27211. Prove that if a ≥ b ≥ c ≥ d ≥ e > 0, n ∈ N then:an

bn+cn + bn

cn+dn + cn

dn+en ≤ an+1

bn+1+en+1 + bn+1

cn+1+dn+1 + cn+1

dn+1+en+1

Daniel Sitaru

PP27212. Prove that if n ∈ N then:(tan 5◦)n

(tan 4◦)n+(tan 3◦)n + (tan 4◦)n

(tan 3◦)n+(tan 2◦)n + (tan 3◦)n

(tan 2◦)n+(tan 1◦)n ≥ 32

Daniel Sitaru

PP27213. Let be a, b, c, d ∈ (0,∞). Prove that:(ab+ cd)2 ≤ (b

5√ab4 + d

5√cd4)(a

5√a4b+ c

5√c4d)

Daniel Sitaru

PP27214. Prove that if a, b ∈ [0, 2] then: a2

b+2 + b3

a+2 + (2− a)b2 ≤ 12

Daniel Sitaru

PP27215. Prove that if a, b ∈ [0, 2] then: a2

b+2 + b3

a+2 + (2− a)b2 ≤ 12

Daniel Sitaru


PP27216. Let be a, b, c, d ∈ (0,∞). Prove that if:(a

3√a2b+ c

3√c2d)(b

3√ab2 + d

3√cd2) ≤ (a2 + c2)(b2 + d2)

Daniel Sitaru

PP27217. Prove that if a, b, c ∈ (0,∞) and:

Δ(a, b) =

��

1 5 10a 4a+ b 6a+ 4ba2 3a2 + 2ab 3a2 + 6ab+ b2

��

then: Δ(a, b) +Δ(b, c) +Δ(c, a) ≤ 0

Daniel Sitaru

PP27218. Prove that if a, b, c > 0 then:(5a+b)(5b+c)(5c+a)

27(a+8c)(b+8a)(c+8b) ≥ 8abc(5a+4b)(5b+4c)(5c+a)

Daniel Sitaru

PP27219. Prove that if a, b, c > 0;q

aba+b +

qbcb+c +

qcac+a = 3

2 then:

a+ b+ c ≥ 32

Daniel Sitaru

PP27220. Prove that if a, b, c > 0, a+ b+ c = 3 then:Pq1 + 1

a2+ 1

(a+1)2≥ 9

12−2(ab+bc+ca) + 3

Daniel Sitaru

PP27221. If a, b, c ≥ 0 then:(a+ 1)a+1 · (b+ 1)b+1 · (c+ 1)c+1 ≤ ea+b+c

√ea2+b2+c2

Daniel Sitaru

PP27222. If x, y, z > 2 then:P

logx−1

�x2+y2

x−1

�> logx 2 + logy 2 + logz 2 + 3

Daniel Sitaru

PP27223. If a, b, c, d > 0 then:P (a+d)2(b2+ac)

b(adb+ac2+cd2)≥ 32

3

Daniel Sitaru


PP27224. If in ABCD tetrahedronAB +DC = AD +BC = AC +BD = 4, RA, RB, RC , RD circumradii in

ΔBCD,ΔACD,ΔABD,ΔABC then: RA +RB +RC +RD ≥ 8√3

3

Daniel Sitaru

PP27225. If 0 < A < B+C, 0 < B < C +A, 0 < C < A+B,A+B+C < πthen: (sinA+ sinB − sinC)(sinB + sinC − sinA)(sinC + sinA− sinB) > 0

Daniel Sitaru

PP27226. Fie matricele A,B ∈ M2 (Q), cu proprietatea AB = BA. Stiindca detA = −20182 si det

�A+B

√2018

�= 0, aratati ca det

�A4 −B4

�este

patrat perfect.

Irimia Alexandra-Valentina

PP27227. Let x, y, z be real number with the property x+ y + z = −1.

Show that:3x+1+3y

34x+1 + 3y + 3z+

3y+1+3z

34y+1 + 3z + 3x+

3z+1+3x

34z+1 + 3x + 3y≥ 1

Mihaela Berindeanu

PP27228. For a, b, c ∈ R+ show that�a2018 − a2016 + 3

� �b2018 − b2016 + 3

� �c2018 − c2016 + 3

�≥

≥�

3√a2 +

3√b2 +

3√c2�3

.

Mihaela Berindeanu

PP27229. Prove thatπR0

cosx cos 2x cos 3x... cos 2017xdx = 0.

Mihaly Bencze

PP27230. ComputeπR0

cosx sin 2x cos 3x sin 4x... cos (2n− 1)x sin 2nxdx.

Mihaly Bencze

PP27231. ComputeπR0

sinx cos 2x sin 3x cos 4x... sin (2n− 1)x cos 2nxdx.

Mihaly Bencze


PP27232. ComputeRarccos

�1−x2

1+x2

�arccos

�4−x2

4+x2

�dx.

Mihaly Bencze


n

�2 ln 2− 1−

1R0

ln (1 + x+ xn) dx

�.

Mihaly Bencze

PP27234. If a1 = 1, an+1 =�an3

�+ 2017 for all n ≥ 1, when [·] denote the

integer part, then compute∞Pk=1

1a2k.

Mihaly Bencze

PP27235. If x1, y1, z1 > 0 and xn+1 = xn

�1 + 1

yn

�, yn+1 = yn

�1 + 1

zn

�,

zn+1 = zn

�1 + 1

xn

�then study the convergence of the sequences

(xn)n≥1 , (yn)n≥1 and (zn)n≥1 .

Mihaly Bencze

PP27236. Let be fk : [0, 1] → R (k = 1, 2, ..., n) continuous functions such

that1R0

f2k (x) dx = 4k3

n(n+1)2, (k = 1, 2, ..., n) . Prove that exist λ ∈ [0, 1] for

whichnP

k=1

fk (λ) ≤ n.

Mihaly Bencze

PP27237. If In =3R2

n−1Pk=1

�x+ k

n

�dx then 5(n−2)

11(5n+1) ≤nP

k=3

1IkIk+1

≤ 4(n−2)9(5n−1) .

Mihaly Bencze

PP27238. We consider xn = min {an, an+1, an+2} andyn = max {an, an+1, an+2} . Prove that if the sequence (an)n≥1 have limit,then the sequences (xn)n≥1 and (yn)n≥1 have limits.

Mihaly Bencze


PP27239. We consider xn+ yn+ zn = xn+1+ yn+1+ zn+1− (n+ 1)xyz = 1.If x, y, z ∈ R then determine all n ∈ N for which x+ y + z = 1.

Mihaly Bencze

PP27240. If ak > 1 (k = 1, 2, ..., n) and (bn)n≥1 is an arithmeticalprogression formed by natural numbers, then

b1 loga1 a2 + b2�loga2 a3

� 1b2 + b3

�loga3 a4

� 1b3 + ...+ bn

�logan a1

� 1bn ≥ n(b1+bn)

2 .

Mihaly Bencze

PP27241. If a, b, c ∈ (0, 1) thenP

loga4b

b+c+3 ≥ 32 .

Mihaly Bencze


k=1

1lg k(k+1) ≥ n

lg(n+1)(2n+1)

3

.

Mihaly Bencze

PP27243. Let A (a) , B (b) , C (c) points in the complex plane, such that(a+ b)3 = (b+ c)3 = (c+ a)3 when a, b, c ∈ C. Prove that the triangle ABCis equilateral.

Mihaly Bencze

PP27244. Solve in Z the equation x3 + y3 + z3 = 3n(2n+1) (x+ y + z) whenn ∈ N.

Mihaly Bencze

PP27245. Computeb−1Pk=1

nk3ab

owhen {·} denote the fractional part and

a, b ∈ N∗.

Mihaly Bencze

PP27246. If f : N → [0, 1) when f (n) =n2017n+

12017

o, {·} denote the

fractional part, then study the injectivity and surjectivity of the givenfunction.

Mihaly Bencze


PP27247. If a, b, c ∈ [0, 1] thenP ab

b+c3+7≤ 1

3 .

Mihaly Bencze

PP27248. If a, b, c > 0 thenqP(a+ 1)4 +

pPb4 +

qP(c− 1)4 ≥

√3�(P

a)2 +P

ab�.

Mihaly Bencze

PP27249. If x, y, z > 0, thench2x (shy + shz)+ ch2y (shz + shx)+ ch2z (shx+ shy) ≤ ch4x+ ch4y+ ch4z.

Mihaly Bencze

PP27250. We consider G =n

2k+12n+1 |k, n ∈ Z

o, H = G× Z and

(x, a) ∗ (y, b) = (xy, a+ b) for all x, y ∈ G and a, b ∈ Z.1). Prove that (H, ∗) is abelian group2). Prove that (H, ∗) ∼= (Q∗, ·)

Mihaly Bencze

PP27251. Determine (an)n≥1 if a1 = 1 andnP

k=1

2k+1akak+1

=√an−1an+1

anfor all

n ≥ 1. ComputenP

k=1

1akak+1

.

Mihaly Bencze

PP27252. We consider the function f : R → R where

f (x) =

�sin

�ax2 + bx+ c+ d

x + ex2

�if x 6= 0

0 if x = 0. Determine all

a, b, c, d, e ∈ R for which f have primitive functions.

Mihaly Bencze

PP27253. Let be K =

A (a, b) =

a 0 −xab (x+ y) b b−ya 0 a

|a, b ∈ R

.

Determine all x, y ∈ R for which (K,+, ·) is a corp.

Mihaly Bencze



1R0

ln�1− x2 + 2xn + x2n

�dx.

Mihaly Bencze

PP27255. If f : [0, 1] → R is a continuous function for whichx1f (x2) + x2f (x3) + ...+ xnf (x1) ≤ 1 for all xk ∈ [0, 1] (= 1, 2, ..., n) then

compute max1R0

f (x) dx.

Mihaly Bencze

PP27256. If f : [0, 1] → R is twice differentiable with f ′′ continuous , then

291R0

(f ′′ (x))2 dx ≥ 105

�6

1R0

xf (x) dx− 2f (1)− f (0)

�2

.

Mihaly Bencze

PP27257. If G =

Ax =

1 ax bx2 − ax0 1 bx0 0 1

|x ∈ R

then determine

all a, b ∈ R for which (G, ·) is abelian group. Prove that (G, ·) ∼= (R,+) .

Mihaly Bencze

PP27258. Compute1R0

�nQ

k=1

{kx}− 1k

�dx, when {·} denote the fractional

part.

Mihaly Bencze

PP27259. Compute In =R xn ln(1+

√1+x2)√

1+x2dx.

Mihaly Bencze

PP27260. If In =1R0

xndxx2+1

then n−112(n+2) ≤

nPk=2

IkIk+1 ≤ n−14n . Compute

limn→∞

nPk=2

IkIk+1Ik+2.

Mihaly Bencze


PP27261. ComputeR �5x5 + 25x4 + 53x3 + 59x2 + 34x+ 8

�·√x3 + 3x2 + 4x+ 2dx.

Mihaly Bencze


k=2

kR1k

arctgxx dx

= π

2 ln (n!) .

Mihaly Bencze

PP27263. Let be M =

��a bc d

�|a, b, c, d ∈ R

�. Denote G the set of

orthogonal matrices from M . Determine all a, b, c, d ∈ R for which(G, ·) isizomorph with Klein’s group.

Mihaly Bencze

PP27264. ComputeR q

x2+2x+2+√x4+4x3+7x2+6x+3

x4+4x3+7x2+6x+3dx.

Mihaly Bencze

PP27265. ComputeR (x2−1)

n−1((x−1)2n−1)((x+1)2n−1)

((x−1)4n+1)((x+1)4n+1)dx, where n ∈ N.

Mihaly Bencze

PP27266. We consider the function

f (x) =

� �sin 1

x

�2016 �cos 1

x

�2017if x 6= 0

c if x = 0. Determine c ∈ R for which the

function f : R → R have primitive function.

Mihaly Bencze

PP27267. Determine all a, b ∈ R for which

4aR0

(1− cos 2x) esinxdx =bR0

(4 + sinx+ sin 3x) esinxdx.

Mihaly Bencze

PP27268. ComputeR

dxsin(ax+b) sin(cx+d) sin(ex+f) sin(gx+h) .

Mihaly Bencze


PP27269. ComputeR x(x2−1)(x2−3)dx

(x2−3x)4+4.

Mihaly Bencze

PP27270. Let be A (a, b, c) =

1 0 0a b c1 0 0

∈ M3 (R) and

G = {A (a, b, c) |a, b, c ∈ Z} . Determine all a, b, c ∈ Z for which (G, ·) isgroup. Prove that (G, ·) ∼= (Z,+) .

Mihaly Bencze

PP27271. If In =

π6R0

(sinx)n cosnxdx then

2n−1 (In−2 + 4In) =sin nπ

6n +

cos(n−1)π

6n−1 for all n ≥ 3.

Mihaly Bencze

PP27272. If a ◦ b =( �

ab

�2n+1if a < 0

(ab)2k+1 if a > 0when a, b ∈ R∗, then determine

all n, k ∈ N for which (R∗, o) is group.

Mihaly Bencze

PP27273. Compute F (x) =R P e−x cos6 xdx

sin7 xif F

�π4

�= 0.

Mihaly Bencze

PP27274. Determine the differentiable function f : (0,+∞) → R such that

f ′ (x) = 2f (x) + 3f(x)x + 5ex for all x > 0 and f (1) = e.

Mihaly Bencze

PP27275. If F : C∗ → C where Fa+b (Z) = (c+ id)Re (z) + (a+ ib) Im (z)for all z ∈ C and G = {Fa+b|a ∈ R, b ∈ R∗} then determine all c, d ∈ R forwhich (G, ◦) is group, when ”o” denote the compozition of functions.

Mihaly Bencze


PP27276. If In =3R2

n−1Pk=1

�x+ k

n

�dx then lim

n→∞Inn = 5

2 . Compute

limn→∞

n�52 − In

n

�.

Mihaly Bencze

PP27277. Compute In =3R2

n−1Pk=2

��x+ k

n

��2dx when [· ] denote the integer

part.

Mihaly Bencze

PP27278. If In =3R2

n−1Pk=1

�x+ k

n

�dx when [· ] denote the integer part then

4(n−2)11(5n+1) ≤

nPk=3

1IkIk+1

≤ 4(n−2)9(5n−1) .

Mihaly Bencze

PP27279. If M =n

k2n+1 |k ∈ {0, 1, 2, ..., 2n}

o, then determine all a, b ∈ R

for which (G, ∗) is abelian group, where x ∗ y = {ax+ by} , {·} denote thefractional part. Prove that (G, ∗) ∼= (Z2n+1, ∗) , n ∈ N∗

Mihaly Bencze and Daniel Sitaru

PP27280. Compute

π2R0

xdx1+sin 2x+sin 4x .

Mihaly Bencze

PP27281. If In =1R0

xndxx2+1

then n2(n+1) ≤

nPk=1

√IkIk+2

k+1 ≤ n+12(n+2) .

Mihaly Bencze

PP27282. If In =1R0

xndxx2+1

then n−112(n+2) ≤

nPk=2

IkIk+1 ≤ n−14n .

Mihaly Bencze


PP27283. Compute1R0

nQk=1

�{kx}− 1

2

�dx, when {·} denote the fractional

part.

Mihaly Bencze

PP27284. Prove that n−14(n+1) <

nPk=2

1R0

xk

x2+1dx < n−1

2n .

Mihaly Bencze

PP27285. If a, b, c > 0 thenP |a− b| ≥ 3(

�

a2−�

ab)2�

a .

Mihaly Bencze

PP27286. If ak > 0 (k = 1, 2, ..., n) thenP

1≤i<j≤n

qaiajai+aj

≤ n−14

nPk=1

ak +n(n−1)

8 .

Mihaly Bencze


1).P |a− b| ≥ 3(s2−3r2−12Rr)

2s

2).P |ra − rb| ≥

3((4R+r)2−3s2)2(4R+r)

3).P ��sin2 A

2 − sin2 B2

�� ≥ 3((4R+r)2−3s2)16R(2R−r)

4).P ��cos2 A

2 − cos2 B2

�� ≥ 3((4R+r)2−3s2)16R(4R+r)

Mihaly Bencze


k=1

(2k + 1) ln (2k + 1) ≤ 2n(n+1)(n+2)3 .

Mihaly Bencze


k=1

k(k+1)2k+1 ≥ 1

2 ln�(2n+1)!2n·n!

�.

Mihaly Bencze

PP27290. In all triangle ABC holds:

1).P 2 sin2 A

2+1

sin2 A2+1

ln�2 sin2 A

2 + 1�≤ 2R−r

R


2).P 2 cos2 A

2+1

cos2 A2+1

ln�2 cos2 A

2 + 1�≤ 4R+r

R

Mihaly Bencze

PP27291. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that

1).nP

k=1

(2Fk + 1) ln (2Fk + 1) ≤ 2p

FnFn+1 (FnFn+1 + 2Fn+2 + n− 2)

2).nP

k=1

(2Lk + 1) ln (2Lk + 1) ≤ 2p(LnLn+1 − 2) (LnLn+1 + 2Ln+2 + n− 8)

Mihaly Bencze

PP27292. If 0 < a ≤ b then1

b−a

�(2b+ 1)2 ln (2b+ 1)− (2a+ 1)2 ln (2a+ 1)

�≤

83

�b2 + ba+ a2

�+ 6 (b+ a) + 2.

Mihaly Bencze

PP27293. In all triangle ABC holds:1).

P 2a+1a+1 ln (2a+ 1) ≤ 4s

2).P 2ra+1

ra+1 ln (2ra + 1) ≤ 2 (4R+ r)

3).P 2ha+1

ha+1 ln (2ha + 1) ≤ s2+r2+4RrR

Mihaly Bencze


k=1

1(2k+1) ln(2k+1) ≥ n

2(n+1) .

Mihaly Bencze

PP27295. If 0 < a ≤ b thenbRa

dx(2x+1) ln(2x+1) ≥ 1

2 lnb(a+1)a(b+1) .

Mihaly Bencze

PP27296. In all triangle ABC holds1). 1− 1

16√sRr

≤ a1+a + b

(1+a)(1+b) +c

(1+a)(1+b)(1+c) ≤ 1− 27(2s+3)3

2). 1− 18s

√r≤ ra

1+ra+ rb

(1+ra)(1+rb)+ rc

(1+ra)(1+rb)(1+rc)≤ 1− 27

(4R+r+3)3

3). 1− R2r ≤ sin2 A

2

1+sin2 A2

+sin2 B

2

(1+sin2 A2 )(1+sin2 B

2 )+

sin2 C2

(1+sin2 A2 )(1+sin2 B

2 )(1+sin2 C2 )

≤≤ 1− 216R3

(8R−r)3


4). 1− R2s ≤ cos2 A

2

1+cos2 A2

+cos2 B

2

(1+cos2 A2 )(1+cos2 B

2 )+

cos2 C2

(1+cos2 A2 )(1+cos2 B

2 )(1+cos2 C2 )

≤≤ 1− 216R3

(10R−r)3

Mihaly Bencze

PP27297. If ak > 0 (k = 1, 2, ..., n), then1− 1

2n

�

n�

k=1ak

≤ a11+a1

+ a2(1+a1)(1+a2)

+...+ an(1+a1)(1+a2)...(1+an)

≤ 1− 1�

1+ 1n

n�

k=1

ak

� .

Mihaly Bencze

PP27298. If f : [a, b] → (0,+∞) is a cpntinuous and increasing function,

then

bRaxαf (x) dx

! bRaxβf (x) dx

! bRaxγf (x) dx

!≥

≥ (bα+1−aα+1)(bβ+1−aβ+1)(α+1)(β+1)(b−a)γ+1

bRaf (x) dx

!γ+2

, when α,β > 0 and γ ≥ 1.

Mihaly Bencze

PP27299. In all triangle ABC holdsParctg

�x sinA

y+x cosA

�+P

arctg�

y sinAx+y cosA

�= π for all x, y > 0.

Mihaly Bencze

PP27300. If a, b > 0; x, y > 0 and ak ∈ [a, b] (k = 1, 2, ..., n) , then

2xynP

k=1

a2k +�x2 + y2

� Pcyclic

a1a2 + n (x+ y)2 ab ≤ (x+ y)2 (a+ b)nP

k=1

ak.

Mihaly Bencze

PP27301. Prove that1R0

cos ((2 [nx] + 1) a) dx = sin 2a2n sin a when [·] denote the

integer part.

Mihaly Bencze and Ovidiu Furdui

PP27302. Solve in Z the equation 2x

(y+z)2+ 2y

(z+x)2+ 2z

(x+y)2= 3

4 .

Mihaly Bencze


PP27303. If 0 < A ≤ B ≤ C ≤ π2 then

(B −A) sinA+ (C −A) sinB + (C −B) sinA ≤ 1.

Mihaly Bencze

PP27304. Prove that1R0

sin ((2 [nx]! + 1) a) dx = sin2 nan sin a when [·] denote the

integer part.

Mihaly Bencze and Ovidiu Furdui

PP27305. Find all triplets of positive numbers (a, b, c) such that

a+ b+ c = 9,a3

bc(b+ c)+

b3

ca(c+ a)+

c3

ab(a+ b)=

3

2.

Jose Luis Dıaz-Barrero

PP27306. Let {an}n≥1 be the sequence defined by a1 = 2, a2 = 7, and for

all n ≥ 3, an = 4an−1 − an−2. For n ≥ 1, show thata2n+1

2is the sum of the

squares of two consecutive integers.


PP27307. Let n ≥ 1 be a positive integer and let z1, . . . , zn be complexnumbers lying in the close left half plane Re(z) ≤ 0. Prove that

1n

nPk=1

k�nk

� h |1−zk|1+|zk|

i2≥ 2n−2

When does equality occurs?



1).P 2+cos A

2

1+cos A2+cos2 A

2

≥ 2 + 4R(8R+s)4R2+2Rs+s2

2).P 2+cos2 A

2

1+cos2 A2+cos4 A

2

≥ 2 +16R2(32R2+s2)16R4+4R2s2+s4

Mihaly Bencze


1).P (r+2ra)ra

r2+rra+r2a≥ (r2+2s2)s2

r4+r2s2+s4+ 2


2).P (r+2ha)ha

r2+rha+h2a≥ 2(Rr+2s2)s2

(Rr)2+2s2Rr+4s4+ 2

Mihaly Bencze


1).P 2+sin A

2

1+sin A2+sin2 A

2

≥ 2 + 4R(8R+r)4R2+2Rr+r2

2).P 2+sin2 A

2

1+sin2 A2+sin4 A

2

≥ 2 +16R2(32R2+r2)16R4+4R2r2+r4

Mihaly Bencze

PP27311. In all acute triangle ABC holdsP 2+cosA

1+cosA+cos2 A≥ 2 +

4R2(s2+8R2−(2R+r)2)(s2−(2R+r)2)

2+4R2(s2−(2R+r)2)+16R4

.

Mihaly Bencze

PP27312. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thennP

k=1

ak+3a3k+a2k+ak+1

≥n�

k=1

ak+3

n�

k=1a3k+

n�

k=1a2k+

n�

k=1ak+1

+ n− 1

Mihaly Bencze

PP27313. If ak ∈ (0, 1] (k = 1, 2, ..., n) then

2nP

k=1

ak+2a2k+ak+1

≥ n+P

cyclic

a1a2+2(a1a2)

2+a1a2+1.

Mihaly Bencze


1).P 2+sinA

1+sinA+sin2 A≥ 2 + (s+2R)R

s2+sR+R2

2).P 2+sin2 A

1+sin2 A+sin4 A≥ 2 +

(s2+2R2)R2

s4+s2R2+R4

Mihaly Bencze

PP27315. In all acute triangle ABC holds1√2

PtgA+

√2P

ctgA+ 2P

cosA+ 2P 1

cosA ≥ 27√2

2

Mihaly Bencze


PP27316. If

ak ∈ (0, 1] (k = 1, 2, ..., n) thennP

k=1

ak+2a2k+ak+1

≤n

�

n

�

n�

k=1ak+2

�

n

�

n�

k=1a2k+

n

�

n�

k=1ak+1

. If ak ≥ 1

(k = 1, 2, ..., n) , then holds the reverse inequality.

Mihaly Bencze

PP27317. If a, b > 0 then�2aba+b +

√ab+ a+b

2 +q

a2+b2

2

��a+b2ab + 1√

ab+ 2

a+b +q

2a2+b2

�≤

≤ 4

r2√2ab

(a+b)√a2+b2

+

r(a+b)

√a2+b2

2√2ab

!2

.

Mihaly Bencze

PP27318. If x ∈�0, π2

�then 1√

2tgx+

√2ctgx+ 2 cosx+ 2

cosx ≥ 9√2

2 .

Mihaly Bencze

PP27319. If x ∈�0, π2

�then�

1√2+ sin 2x

sinx+cosx

� �√2 + sinx+cosx

sin 2x

�≤ 2(1+sin 2x)

sin 2x .

Mihaly Bencze

PP27320. If ak > 0 (k = 1, 2, ..., n) , then

Pcyclic

�2a1a2a1+a2

+

qa21+a22

2

� a1+a2

2 +

r2a21a

22

a21+a22

!≤ 2

nPk=1

ak.

Mihaly Bencze

PP27321. Prove that2n(n+1)P

k=1

√2k−1+2

√2k+

√2k+1√

4k2−2k+2k+√4k2−1+

√4k2+2k

= 2n.

Mihaly Bencze

PP27322. If ak =√2k +

√2k + 1 and bk =

√2k − 1 +

√2k (k = 1, 2, ..., n)

then

�nP

k=1

ak

��nP

k=1

1ak

�+ n

nPk=1

bk ≤ n+nP

k=1

ak +

�nP

k=1

bk

��nP

k=1

1bk

�.

Mihaly Bencze



k=1

��nk

�!�2 ≥

nPk=1

nknk.

Mihaly Bencze

PP27324. Determine all n ∈ N∗ for whichnP

k=1

k! ≥s

nnP

k=1

kk.

Mihaly Bencze

PP27325. Prove that 2nP

k=1

ln k ≥ (n+ 1) ln (n+ 1) .

Mihaly Bencze

PP27326. If ni ∈ N∗ (i = 1, 2, ..., k) then

2k�

i=1ln(ni!)

k�

i=1ni

k�

i=1ni

≥nQ

i=1(lnni)

ni .

Mihaly Bencze

PP27327. Determine all ni ∈ N∗ (i = 1, 2, ..., k) for where�kP

i=1ni!

�2

≥ knP

i=1nnii .

Mihaly Bencze

PP27328. Prove that

1).nP

k=1

(Lk+2 − 2Lk − 3)2 ≥ 2 (Ln+2 − 3)2 − 1

2).nP

k=1

�LnLn+1 − 2L2

k − 2�2 ≥ 2 (LnLn+1 − 2)2 − 1

Mihaly Bencze

PP27329. Prove that

1).nP

k=1

�n(n+1)

2 − 2k�2

≥ n2(n+1)2

2 − 1

2).nP

k=1

�n(n+1)(2n+1)

6 − 2k2�2

≥ n2(n+1)2(2n+1)2

18 − 1 for all n ≥ 4

Mihaly Bencze



1).P

sin2A ≤P�

sin2 A1+sin2 A

�cos2 A

2).P

cos2A ≤P�

cos2 A1+cos2 A

�sin2 A

Mihaly Bencze

PP27331. Prove that

1).nP

k=1

(Fn+2 − 2Fk − 1)2 ≥ 2 (Fn+2 − 1)2 − 1

2).nP

k=1

�FnFn+1 − 2F 2

k

�2 ≥ 2F 2nF

2n+1 − 1 for all n ≥ 4

Mihaly Bencze

PP27332. In all triangle ABC holds 3 +P sinA

sinB ≥P�1+sinAsinA

�sinB

Mihaly Bencze

PP27333. In all acute triangle ABC holds 3 +P cosA

cosB ≥P�1+cosAcosA

�cosB.

Mihaly Bencze

PP27334. In all acute triangle ABC holds

1). 3 +P

tgA ≥P�1+cosAcosA

�sinA

2). 3 +P

ctgA ≥P�1+sinAsinA

�cosA

Mihaly Bencze

PP27335. If ak > 0 (k = 1, 2, ..., n) , thenP a21(a1+2a2+...+nan)(a2+a3+...+an)

2 ≥ 2n

(n−1)2(n+1)

�

n�

k=1

ak

� .

Mihaly Bencze

PP27336. If x ∈ R then (sinx)2 cos2 x + (cosx)2 sin

2 x ≥ 1.

Mihaly Bencze

PP27337. If a, b ≥ 1 then b2k−1

(1+a)1b+ a2k−1

(1+b)1a≥ (a+b)2k−1

22k−1 for all k ∈ N∗.

Mihaly Bencze


PP27338. If ak ≥ 1 (k = 1, 2, ..., n) and p ∈ N∗ thenP

cyclic

a2p−12

(1+a1)1a2

≥ 12n2p−2

�nP

k=1

ak

�2p−1

.

Mihaly Bencze

PP27339. We have the following inequalities:

1). 49nP

k=1

ln 7Fk5+2Fk

+ 45nP

k=1

Fk lnFk ≥ 720 (Fn+2 − n− 1)

2). 49nP

k=1

ln 7Lk5+2Lk

+ 45nP

k=1

Lk lnLk ≥ 720 (Ln+2 − n− 3)

3). 49nP

k=1

ln7F 2

k

5+2F 2k+ 90

nPk=1

F 2k lnFk ≥ 720 (FnFn+1 − n)

4). 49nP

k=1

ln7L2

k

5+2L2k+ 90

nPk=1

L2k lnLk ≥ 720 (LnLn+1 − n− 2)

Mihaly Bencze

PP27340. If 0 < a < b, f : [a, b] → (0,+∞) is a continuous function such

that nxnRaf (x) dx =

bRxn

f (x) dx, then compute limn→∞

n (a− xn) .

Mihaly Bencze

PP27341. ComputeRthx ln (1 + chx) dx.

Mihaly Bencze

PP27342. Prove that 3n(n−1)2 <

nPk=1

�k2 − k + 1

�sin π

k < πn(n−1)2 .

Mihaly Bencze


k=2

1R0

xk2−k−1dx1+x4k+1 ≥ 2(n−1)(n+3)

3(n+2) .

Mihaly Bencze

PP27344. If n�

3√xn + 5

√xn

�= 2n+ 1 for all n ≥ 1 then compute

limn→∞

n�158 − n+ nxn

�.

Mihaly Bencze


PP27345. If x1 = 2 and lnxn+1 = 2�xn−1xn+1

�for all n ≥ 1, then compute

limn→∞

n (1− xn) .

Mihaly Bencze

PP27346. If x1 = 1 and n2x2n +�n2 − 1

�x2n−1 ≤

�2n2 − 1

�xnxn−1 for all

n ≥ 2, then

�nP

k=2

xkxk−1

�= n− 2 when [·] denote the integer part.

Mihaly Bencze

PP27347. If a1, b1 ∈ R and 2n2an+1 = n2a2n − b2n andnbn+1 = − (n+ 1) anbn for all n ≥ 1 then compute lim

n→∞anbn.

Mihaly Bencze

PP27348. If a > 1, b ∈ R and f : R → R such that f (ax+ b) = f (x) for all

x ∈ R then f (x) = f�

b1−a

�.

Mihaly Bencze

PP27349. If V (a1, a2, ..., an) =

��

1 1 ... 1a1 a2 ... an− − −− −−an−11 an−1

2 −− an=1n

��then

V 2 (a1, a2, ..., an−1, an+1)+

+V 2 (a1, a2, ..., an−2, an+1, an)+ ...+V 2 (an+1, a2, ..., an) ≥ 1nV

2 (a1, a2, ..., an)for all ak ∈ R (k = 1, 2, ..., n.)

Mihaly Bencze

PP27350. If A =

�0 12 0

�, then determine all X ∈ M2 (z) such that

AX = X3A.

Mihaly Bencze

PP27351. If x1 = 1 and xn+1 =xnn + n+1

n2 for all n ≥ 1 then computelimn→∞

n (1− nxn) .

Mihaly Bencze



�F1

n+L1+ F2

n+L2+ ...+ Fn

n+Ln

�when Fk, Lk denote

the kth Fibonacci respective Lucas number.

Mihaly Bencze

PP27353. If a1 = 1 and a2n+1 = an + nn+1 for all n ≥ 1 then compute

limn→∞

n�1+

√5

2 − an

�.

Mihaly Bencze

PP27354. If A ∈ Mn (R) such that A2 = On then determine all matricesX ∈ Mn (R) for which det

�XA+X2

�≥ 0 ≥ det

�XA−X2

�.

Mihaly Bencze

PP27355. If (n− 1)xn+1 = n2xn for all n ≥ 2 and x1 = 1 then compute

limn→∞

n

�e− 2−

nPk=2

k−1kxk

�.

Mihaly Bencze

PP27356. If a1 = 1, an+1 (1 + nan) = 1 for all n ≥ 1 then compute

limn→∞

�1a21

+ 2a22

+ ...+ na2n

�.

Mihaly Bencze

PP27357. If x1 =1√2; 2x2n+1 = 1 + xn for all n ≥ 1 then compute

limn→∞

n�2π − x1x2...xn

�

Mihaly Bencze

PP27358. Let A,B ∈ M2 (C) such that AB =

�a ba+ 1 b

�. Determine all

a, b ∈ C for which BA− (BA)−1 = (a+ 1) I2.

Mihaly Bencze

PP27359. If f : [0,+∞) → R is a function for which limxց0

f (x) exist and is

finite and f (x+ y + z) = f�√

xy +√yz +

√zx

�for all x, y, z > 0 then f is

constant.

Mihaly Bencze


PP27360. Prove that��

k! (k + 1)! (k + 2)! ... (k + n− 1)!(k + 1)! (k + 1)! (k + 2)! ... (k + n− 1)!(k + 2)! (k + 2)! (k + 2)! ... (k + n− 1)!−−− −−− −−− − −−−−−(k + n− 1)! (k + n− 1)! (k + n− 1)! ... (k + n− 1)!

��

is divisible by

1!2!...n! for all k, n ∈ N∗.

Mihaly Bencze

PP27361. Prove that

��

�x2 + 1

�2(xy + 1)2 (xz + 1)2

(xy + 1)2�y2 + 1

�2(yz + 1)2

(xz + 1)1 (yz + 1)2�z2 + 1

�2

��≤

≤ 23

�2P

x6 + 15P

x2y2�x2 + y2

�− 6

Pxy

�x4 + y4

�− 20

Px3y3

�.

Mihaly Bencze

PP27362. If A,B ∈ M2 (C) thendet

�A2 +B2 +AB +BA

�· det

�A2 −B2 +AB −BA

�=

= det�A4 −B4 +A3B −B3A+ABA2 −BAB2 + (AB)2 − (BA)2

�if and

only if�A2B2 −B2A2 +A2BA−B2AB +AB3 −BA3 +AB2A−BA2B

�2= O2.

Mihaly Bencze

PP27363. If x1, y1 > 0 and xn+1 = xn + 2yn; yn+1 = xn + yn, cn = anbn; then

compute limn→∞

n�√

2− cn�.

Mihaly Bencze

PP27364. Prove thatn(n+3)

2(n+1)(n+2) <nP

k=1

√2k(2k+1)−

√(2k−1)(2k+2)

k(k+2) < n(3n+5)4(n+1)(n+2) .

Mihaly Bencze

PP27365. If x1 ∈ (2, 3) and xn+1 + 4xn = x2n + 6, then computelimn→∞

n (2− xn) .

Mihaly Bencze


PP27366. Compute limx→0

�nP

k=0

xk��

nPk=0

1xk

�, when [·] denote the integer

part.

Mihaly Bencze

PP27367. If√nxn + n+

√nxn ≤ 2

√n ≤ √

nxn + n+ 1 +√nxn + 1 the

compute limn→∞

n�xn − 9

16

�.

Mihaly Bencze

PP27368. If x1 = 3, xn+1 = xn + 2n+ 3 for all n ≥ 1 then compute

limn→∞

n

�12 − n

�ln 2 + ln

�nP

k=1

1xk

��.

Mihaly Bencze

PP27369. ComputeR (4x4+1)dx

x(x4+1)(x8+x4+1)(x16+2x12+x8+2).

Mihaly Bencze

PP27370. ComputeR cos(x−π

4 ) sin(π4−x)dx

(ex+cosx)(ex−sinx) .

Mihaly Bencze

PP27371. ComputeR (x2+1)(sinx+cosx)−x(sinx−cosx)+2

(x2+2 sinx+1)2dx.

Mihaly Bencze

PP27372. ComputeR 1+ min

x≤t≤x+1(t2−2t)

1+ minx≤t≤x+1

(t2−3t)dx.

Mihaly Bencze

PP27373. Determine all twice differentiable f, g : (0,+∞) → R functions

for which f (1) = 2, g (1) = −1 and f(x)x is a primitive for g (x) and g(x)

x is aprimitive for f (x) . Prove that f ′′ (x) + g′′ (x) =

�x2 + 3

�(f (x) + g (x)) for

all x > 0.

Mihaly Bencze


PP27374. If a, b > 0, then

1).1R0

xadx1−xb ≥ 1

a+1 + 1a+b+1 + ...+ 1

a+nb+1

2).1R0

xadx1+xb ≤ 1

a+1 − 1a+b+1 +

1a+2b+1 + ...− 1

a+(2n−1)b+1 +1

a+2nb+1 for all n ∈ N∗

Mihaly Bencze

PP27375. Compute In =

12R

− 12

(arcsin)2ndx√1−x2+1+x

.

Mihaly Bencze

PP27376. Compute In =Rarcsinn (sinx) dx.

Mihaly Bencze

PP27377. Determine all a, b ∈ R for which F (0) = π4 , F (−1) = −π

2 when

F (x) =R (ax+b)dx

x(x+1)(x+a)(x+b)+a .

Mihaly Bencze

PP27378. ComputeR �

x(2x+1)2

sinx+ x+1(2x+1)2

cosx�dx.

Mihaly Bencze

PP27379. ComputeR �arctgx

x

�2dx.

Mihaly Bencze

PP27380. ComputeR

dx1+

√x+

√x+1+

√x+2

.

Mihaly Bencze

PP27381. Computex(3(7+4x2+4

√3+4x2)−4(1+

√2+3x2) ln(1+

√2+3x2))dx

(1+√2+3x2)(1+

√3+4x2)(7+4x2+4

√3+4x2)

.

Mihaly Bencze


PP27382. Compute1R0

�{2x}− 1

3

� �{3x}− 1

2

�dx where {·} denote the

fractional part.

Mihaly Bencze

PP27383. Prove that the function f : R → R where f5 (x) = 5f(x) = x forall x ∈ R is primitivable.

Mihaly Bencze

PP27384. Determine all a, b ∈ R∗ for

limn→∞

1na

nPk=1

√na − ka = lim

n→∞

nPk=1

nb−1

nb+kb= π

ab .

Mihaly Bencze

PP27385. a). Find the maximal length of a stick that can be turnedaround in a unit square! By stick we mean a segment with endpoints A,B.This segment is inside or on the boundary of the square of side length 1. Byturning around we mean that the stick can not be broken or bent, it can notbe lifted out of the plane of the square, and no point of it can ever be outsideof the square in the course of its morion. the stick can slide or rotate, and inits final position, the endpoint A must be where B was initially, and B mustbe where A was initially. try to prove your assertion, not just guess it.b). Find the maximal length of a stick that can be turned around in a unitcube!

Nicolas Martin

PP27386. a). Find all one to one functions f : R → R that satisfy theequation af (f (x)) + bf (x) + c = 0 for all x ∈ R, b 6= 0b). Show that if we assume that f above is onto, then f is one to onec). Find all functions f : R → R such that f (f (x)) = K for all x ∈ R (a, b, cand K are given real constants)

Nicolas Martin

PP27387. a). Consider two sets of two real numbers each: X = {a, b} andY = {α,β} . Show that the necessary and suffiecient condition that whenplotted on the real line, the elements of each set have precisely one elementof the other set between them (call this situation intertwine) is:(α− a) (α− b) (β − a) (β − b) < 0


b). Assuming that P (x) = ax2 + bx+ c and Q (x) = Ax2 +Bx+C have tworeal roots each, show that the necessary and sufficient condition that theyintertwine is that: (aC − cA)2 < (aB − bA) (bC − cB) .

Nicolas Martin

PP27388. If xk > 0 (k = 1, 2, ..., n) and a, b > 0 such thatnP

k=1

xak = n, then

Pcyclic

xa+b1

xa2+xa

3+...+xan−1

≥ nn−1 .

Mihaly Bencze

PP27389. Given the suite of numbers (an) , n ∈ N defined by a1 =13 and

54an+1 + 1 = 12�√

6an + 2 + 2an�, calculate the number pattern and

limn→∞

an.

Mihaela Berindeanu

PP27390. Let C ∈ Mn (C) with the property X10 +X + 2In = On. Showthat X2 −X + In is invertable.

Irimia Alexandra-Valentina

PP27391. Let A,B be two matrix A,B ∈ M2017 (R) with the propertyA2 +B2 =

√3AB. Show that (AB −BA) = 0.

Mihaela Berindeanu

PP27392. Solve the equation:n

[x]x+1

o+h{x}x+1

i= 2

3 , where [·] denote the

integer part and {·} denote the fractional part.

Pirkuliyev Rovsen

PP27393. Compute:R c2017−2016(x2016+x2015)−2 sinx

x2017+x2016−sinx−cosxdx. Generalization:

R xn−(n−1)(xn−1+xn−2)−2 sinx

xn+xn−1−sinx−cosxdx, where n ∈ N, n ≥ 2.

Pirkuliyev Rovsen

PP27394. Solve in N the equationn

1√x

o+n

1√y

o+n

1√z

o= 1, where {·}

denote the fractional part.

Pirkuliyev Rovsen


PP27395. Compute L = limn→∞

200nP

k=1+100m

1k +

400nPk=1+200m

1k +

400nPk=1+300m

1k

!.

Daniel Sitaru

PP27396. Compute L = limx→∞

n�

i=1(x+ai)

x+ai

�

x+n�

i=1ai

�nx+n�

i=1ai

where n ∈ N∗;

a1, a2, ..., an ∈ (0,∞) are fixed.

Daniel Sitaru

PP27397. Compute L = limn→∞

�nP

k=1

1(2k)2

·nP

k=1

1(2k+1)2

·nP

k=1

1(2k+2)2

�.

Daniel Sitaru

PP27398. Prove that if x, y, z ∈ (0,∞) ; x+ y + z = 1 thenPcyc

xy�2

1x − 1

�> ln 2.

Daniel Sitaru

PP27399. Prove that limn→∞

nPk=1

cos 1k ≥ 1− π2

12 .

Daniel Sitaru

PP27400. In all triangle ABC holdsP (a+b)2−c2

(2(a2+b2)−c2)rc≤ 1

r .

Mihaly Bencze

PP27401. In all triangle ABC holdsP ((a+b)2−c2) sin4 C

2

2(a2+b2)−c2≤ 8R2+r2−s2

8R2 .

Mihaly Bencze

PP27402. Let ABC be a triangle. Determine all λ ≥ 1 for whichP �λtgA

2 − ctgA�≤ (λ− 1)

√λ.

Mihaly Bencze


PP27403. If xk ∈ R (k = 1, 2, ..., n) andnP

k=1

xk = 0 then

nPk=1

log3�1 + axk + a2xk

�≥ n for all a > 0.

Mihaly Bencze

PP27404. Determine all functions f : C → C such that (f ◦ f) (x) = x2018

for all x ∈ C.

Mihaly Bencze

PP27405. Prove that∞Pk=2

log2e

�

23√

3...k√k

> e− 32 .

Mihaly Bencze


x51 − 5x41 − 20x31 = 40x22 + 40x3 + 17x52 − 5x42 − 20x32 = 40x23 + 40x4 + 17−−−−−−−−−−−−−−−−−−x5n − 5x4n − 20x3n = 40x21 + 40x2 + 17

.

Mihaly Bencze

PP27407. Solve the equation(x+ 2017)2018 + 2017 = 2018

√x+ 20172018 + 20172018.

Mihaly Bencze

PP27408. Let be ak > 0 (k = 1, 2, ..., n) in arithmetical progression. Prove

thatnP

k=1

1(ak+ak+1)

2 ·n+1Pk=2

1(ak+ak+1)

2 ≥ n2

(a1+a2)2(an+1+an+2)

2 .

Mihaly Bencze

PP27409. Prove that n√n+ 1

n <�1 + 1

n

�n√n for all n ≥ 2.

Mihaly Bencze

PP27410. Solve in Z the equationx (x+ 1) (x+ 2) (x+ 3) (x+ 4) (x+ 5) = y5 + 1687.

Mihaly Bencze


PP27411. In all triangle ABC holdsP (rb+rc)rarc

�

(r2a+r2b)(r2a+r2c )≤ 4R+ r.

Mihaly Bencze

PP27412. In all triangle ABC holds�s2 − 3r2 − 12Rr

� �s2 − 2r2 − 8Rr

�2 ≥P (a− b)2 .

Mihaly Bencze

PP27413. In all triangle ABC holds�(4R+ r)2 − 3s2

��(4R+ r)2 − 2s2

�2≥P (ra − rb)

2 .

Mihaly Bencze


� �8R2 + r2 − s2

�2 ≥ 1024R6P �

sin2 A2 − sin2 B

2

�2.

Mihaly Bencze

PP27415. ComputeR (x+1)(x+3)dx

x4+12x3+48x2+84x+57.

Mihaly Bencze

PP27416. In all triangle ABC holds4�s2 − 3r2 − 12Rr

� �s2 − r2 − 4Rr

�2 ≥P (a− b)2 .

Mihaly Bencze

PP27417. In all triangle ABC holdsP ra

rb≤ (4R+r)((4R+r)2−2s2)

3s2r.

Mihaly Bencze

PP27418. In all triangle ABC holdsP�

sin A2

sin B2

�2

≤ (2R−r)(8R2+r2−s2)3Rr2

.

Mihaly Bencze


��(4R+ r)2 − s2

�2≥ 1024R6

P �cos2 A

2 − cos2 B2

�2.

Mihaly Bencze


PP27420. In all triangle ABC holdsP −a+b+c

a−b+c ≤ s2−2r2−8Rr3r2

.

Mihaly Bencze

PP27421. If xk ∈ (0, 1) (k = 1, 2, ..., n) , thennP

k=1

arctgxk ≥

�

n�

k=1xk

�2

n+n�

k=1x2k

.

Mihaly Bencze

PP27422. 1). If xk ∈ (0, 1) (k =, 2, ..., n) , thennP

k=1

tg

�πx2

k

2(x2k+1)

�<

nPk=1

xk

2). If xk > 1 (k = 1, 2, ..., n) thennP

k=1

tg

�πx2

k

2(x2k+1)

�>

nPk=1

xk.

Mihaly Bencze

PP27423. In all triangle ABC holdsP�

cos A2

cos B2

�2

≤ (4R+r)((4R+r)2−s2)3Rs2

.

Mihaly Bencze

PP27424. Determine all x ∈ R\Q for which x (x+ 1) (x+ 2) ... (x+ n) andxn+1 (x+ n+ 1) are integer numbers, for all n ∈ N.

Mihaly Bencze

PP27425. In all triangle ABC holdsP

tg�

π2(π2+A2)

�≥ 9.

Mihaly Bencze

PP27426. In all triangle ABC holds maxnP

tg πr2

2(r2+r2a);P

tg πr2

2(r2+h2a)

o≤ 1.

Mihaly Bencze

PP27427. In all tetrahedron ABCD holdmax

nPtg πr2

2(r2+h2a);P

tg 2πr2

4r2+r2a

o≤ 1.

Mihaly Bencze


k=1

tg

�π

2(1+k2(k+1)2)

�≤ n

n+1 .

Mihaly Bencze

matsefi/octogon/volumes/octogon_2017_2_proposed... · proposed problems 475 proposed problems...

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