octogon mathematical magazine, vol. 19, no.2, october ...432 octogon mathematical magazine, vol. 19,...

430 Octogon Mathematical Magazine, Vol. 19, No.2, October 2011

Proposed problems

PP. 18690. 20If x, y, z > 0, then∑ (√x+√z)√x+y√

xz≥ 6

√2.

Mihály Bencze

PP. 18691. If xk > 0 (k = 1, 2, ..., n) , thenn∑k=1

(n− k + 1) ak −n∑k=2

k√a1a2...ak ≥ (n− 1) max

1≤i 0, thenx2

ax+by+cz +y2

ay+bz+cx +z2

az+bx+cy ≥x+y+za+b+c .

Mihály Bencze

PP. 18693. If xk > 0 (k = 1, 2, ..., n) , then∑√x21 + x

22 ≤

√2

n∑k=1

xk +12

∑(x21 − x22

).

Mihály Bencze

PP. 18694. If x, y, z, a, b, c > 0, thena

ax+by+cz +b

bx+cy+az +c

cx+ay+bz ≥3∑ab

(x+y+z)∑a2.

Mihály Bencze

PP. 18695. If x, y, z, t > 0, then∑ 1

23x4y+7y4z+11z4t+10t4x≤

∑xyz

51x2y2z2t2.

Mihály Bencze

PP. 18696. If x, y, z > 0, then:

1). (∑x)2

6∑x2

≤∑ x2

3x2+2y2+z2≤ 118

∑ y+2zx

2).∑ 1

3x2+2y2+z2≤ 16

∑ 1xy

Mihály Bencze

20Solution should be mailed to editor until 30.12.2013. No problem is ever permanentlyclosed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

Proposed Problems 431

PP. 18697. If x, y, z > 0, then:

1). (∑x)2∑

x2+∑xy

≤∑ x2

x2+yz≤ 14

∑ y+zx

2).∑ 1

x2+yz≤ 12

∑ 1xy

Mihály Bencze

PP. 18698. If x, y, z > 0, then 4∑ x2y

z +∑ y3

x + 2∑z2 ≥ 7

∑xy.

Mihály Bencze

PP. 18699. If x, y, z > 0, then 4∑x2z +

∑ y2z +

∑ z3x2

≥ 6∑ xz

y .

Mihály Bencze

PP. 18700. If x, y, z > 0, then∑ 1

4x3y+y3z+2z3x≤

∑xy

7x2y2z2.

Mihály Bencze

PP. 18701. In all triangle ABC holds 32r ≥∑ 1

ha−r ≥3R .

Mihály Bencze

PP. 18702. In all triangle ABC holds∑ 1

a3+bc≤ 4s+3108sRr .

Mihály Bencze

PP. 18703. Solve in R the following system:1x1

+ 1√1−x22

= 1x2 +1√1−x23

= ... = 1xn +1√1−x21

= 2√2.

Mihály Bencze

PP. 18704. Prove thatk∑p=1

(−1)k−1(k−1p−1)n+k =

(k−1)!(n+1)(n+2)...(n+k) .

Mihály Bencze

PP. 18705. In all triangle ABC holds∏r

1hb

+ 1hc

a ≤ (R+ r)2r .

Mihály Bencze

PP. 18706. In all triangle ABC holds(∑ a2+b2

c

)(∑ a+bwa+hb

)≥ 8

√3s.

Mihály Bencze


PP. 18707. If x, y, z are the solution of the Diophantine equation

x2 + y2 = z2, then x3 + y3 + z3 is a composed number.

Mihály Bencze

PP. 18708. If x0 =52 and xn+1 = x

3n − 6x2n + 12xn − 6 for all n ≥ 0, then

8 (xn − 2) = (x1 − 2)2 (x2 − 2)2 ... (xn−1 − 2)2 for all n ∈ N∗.

Mihály Bencze

PP. 18709. If x, y, z > 0 then(∑

xn+3∑xn+2

)2+(∑

xn+1∑xn

)2≥∑xn+3∑xn+1

+∑xn+1∑xn−1

for all n ∈ N∗.

Mihály Bencze

PP. 18710. Determine all b ≥ 2, b ∈ N and all n ∈ N for which200...00︸︷︷︸n−time

11(b) is a perfect square

Mihály Bencze and Béla Kovács

PP. 18711. Prove that there exist infinitely many x, y, z different positiverational numbers such that x2n+1 + y2n + z2n−1, y2n+1 + z2n + x2n−1,z2n+1 + x2n + y2n−1 are squares of rational numbers for all n ∈ N.

Mihály Bencze

PP. 18712. Prove that 25F 2n − 4L2n = 5L2n+1 − 4Ln−1Ln+1 for all n ∈ N∗,where Fn and Ln denote the n

th Fibonacci respective Lucas numbers.

Mihály Bencze

PP. 18713. Prove thatn∏k=0

(2kk

)(2n−2kn−k

)≤(

4n

n+1

)n+1.

Mihály Bencze

PP. 18714. Prove that the equation1320x3 = (y1 + y2 + y3 + y4) (z1 + z2 + z3 + z4) (t1 + t2 + t3 + t4 + t5) haveinfinitely many solutions in Fibonacci numbers set.

Mihály Bencze


PP. 18715. Prove that the equation x3+3y3+ z3 = 3t3+6u3 have infinitelymany solutions in the set of Fibonacci numbers. Solve the equation in N.

Mihály Bencze

PP. 18716. Determine all k, n ∈ N for which 5n − 1 is divisible by 2k.

Mihály Bencze

PP. 18717. If F1 = 1, F2 = 1 and Fn+2 = Fn+1 + Fn for all n ∈ N, thendetermine all p ∈ N for which Fpn is divisible by p− 1.

Mihály Bencze

PP. 18718. If an+1 = a2n + 5an + 1 for all n ∈ N , then determine all

a0, k, p ∈ N for which ak = 2011p.

Mihály Bencze

PP. 18719. If xn =√24

((1 +

√2)n − (1−√2)n), then compute

∞∑n=1

1(xn;xn+1)

2 .

Mihály Bencze

PP. 18720. Let ak, bk ∈ N (k = 1, 2, 3) such that a1n+ b1 and a2n+ b2 areperfect squares for n ∈ N . Determine all ak, bk ∈ N (k = 1, 2, 3) for whicha3n+ b3 is not prime numbers.

Mihály Bencze

PP. 18721. Solve the following system:logx+3

(y2 − 9y − 10

)≤ logz+3 12

logy+3(z2 − 9z − 10

)≤ logx+3 12

logz+3(x2 − 9x− 10

)≤ logy+3 12

.

Mihály Bencze

PP. 18722. Solve the following system:

2− x =

(2− y3

)32− y =

(2− z3

)32− z =

(2− x3

)3 .Mihály Bencze


PP. 18723. If p1 = 2 and pn+1 is biggest prime divisor of 1 + p1p2...pn, thenexist k ∈ N such that pk = 2011?

Mihály Bencze

PP. 18724. If x > e then compute∫ 1+2x2 lnx+2x2(x2−2) ln2 x+3x4 ln4 x

(1−x2 ln2 x)2 dx.

Mihály Bencze

PP. 18725. Solve the equation6(x2 + 2x+ 2

)√5− x =

(x2 + 2x+ 12

)√11x− 5.

Mihály Bencze


2√2x+ 6 = y2 − 6

2√2y + 6 = z2 − 6

2√2z + 6 = x2 − 6

.

Mihály Bencze

PP. 18727. If x > 0 then compute∫ 1+x+2x(x+1) lnx+3x2 ln2 x

x(1+x lnx+x2 ln2 x)dx.

Mihály Bencze

PP. 18728. Solve in Z the equation x3 + y3 + z3 = 9xyz + 27.

Mihály Bencze

PP. 18729. If a, b > 0, then(b3a +

ba+2b +

4a2a+b

)(a3b +

ab+2a +

4b2b+a

)≥ 4

Mihály Bencze

PP. 18730. In all triangle ABC (A 6= B 6= C) we have∑( sin A

2

sin B−C2

)2≥ 2.

Mihály Bencze

PP. 18731. If a, b > 0, then√3(a+b)2a+b +

b(a+2b)a2+ab+b2

+√

3(a+b)2b+a +

a(b+2a)a2+ab+b2

≥ 3( √

a2a+b +

√b

2b+a

).

Mihály Bencze


PP. 18732. If a, b > 0, then 2a3(2a+b) +2b

3(2b+a) +(

a2a+b

)2+(

b2b+a

)2≤ 23 .

Mihály Bencze

PP. 18733. If a, b > 0, then(a2+b2)

2

ab(a+2b)(b+2a) ≥(

a2a+b +

b2

a2+ab+b2

)(b

2b+a +a2

b2+ba+a2

).

Mihály Bencze

PP. 18734. If a, b > 0, then(

a+b5a+4b +

a8a+b

)(a+b

5b+4a +b

8b+a

)≤ 19 .

Mihály Bencze

PP. 18735. If a, b > 0, then√

13 +

2a2

a2+ab+b2+√

13 +

2b2

a2+ab+b2≥ 3(a

2+4ab+b2)(2a+b)(2b+a) .

Mihály Bencze

PP. 18736. If ak > 0 (k = 1, 2, ..., n) , then∑cyclic

√13 +

a21+a22

a21+a1a2+a22≥ n.

Mihály Bencze

PP. 18737. If a, b > 0, then4(3a4+2a3b+2a2b2+2ab3+3b4)(3a2+2ab+b2)(3b2+2ba+a2)

+ (2a−b)2

2a2+b2+ (2b−a)

2

2b2+a2≥ 2.

Mihály Bencze

PP. 18738. If a, b > 0, then(4ab3a+b +

a2

a+b

)(4ab3b+a +

b2

a+b

)≤ 14 (2a+ b) (2b+ a) .

Mihály Bencze

PP. 18739. If x ∈ R, then 48+sin2 x cos4 x

+ 12+3 cos2 x

+ 24+sin2 x

≥ 43 .

Mihály Bencze

PP. 18740. If a, b > 0, then2a(3b2−a2)

(a+b)(√2a+b)

+2b(3a2−b2)

(a+b)(√2b+a)

+(√

2− 1)(a+ b) ≤ 3√

2a+b+ 3√

2b+a.

Mihály Bencze


PP. 18741. If a, b, c, d > 0 and ab+ cd = 2, then1

1+2a +1

1+2b +1

1+2c +1

1+2d ≥12

(2+ab)(2+cd) .

Mihály Bencze

PP. 18742. If a, b, c, d > 0 and a+ b+ c+ d = 2, then√a+2ba2+2b2

+√

b+2ab2+2a2

+√

c+2dc2+2d2

+√

d+2cd2+2c2

≤ 8√(a+b)(c+d)

.

Mihály Bencze

PP. 18743. If a, b, c, d > 0 and a+ b+ c+ d = 4 then√4a2+ab+4b2

(a2+2b2)(b2+2a2)+√

4c2+cd+4d2

(c2+2d2)(d2+2c2)≤ 4(a+b)(c+d) .

Mihály Bencze

PP. 18744. If a, b ∈(0, π2

)and a ≤ b then

2√2 ln

(tga+3+2√2)(tgb+3−2

√2)

(tga+3−2√2)(tgb+3+2

√2)

≤ 16b∫a

dx(sinx+cosx)4

≤ sin(2b−2a)sin 2a sin 2b + 2(b− a).

Mihály Bencze

PP. 18745. If a, b > 0, then4(a2+b2)(2a2+ab+2b2)(3a2+ab+b2)(3b2+ba+a2)

+ a2+4ab+b2

(2a+b)(a+2b) ≥ 2.

Mihály Bencze

PP. 18746. If a, b > 0, then13 ≤

a4+2a3b+8a2b2+2ab3+b4

(4a2+2ab+b2)(4b2+2ba+a2)+ 2a

4+3a2b2+2b4

(4a2+b2)(4b2+a2)≤ 12 .

Mihály Bencze

PP. 18747. If the equation x2 −mx+m = 0 have real roots, then20m2 + 158 ≥ 80m+ 79 |m− 2| for all m ∈ R.

Mihály Bencze

PP. 18748. Let be f : R→ R where f (x) = x− a [x] + [bx] where [·] denotethe integer part, and a, b ∈ Z. Prove that the points M (a, b) lie on the linex− y − 2 = 0 if and only if (f ◦ f) (x) = x for all x ∈ R.

Mihály Bencze


PP. 18749. If ak > 0 (k = 1, 2, ..., n), then

1).∑ a1

min{a2,a3,...,an} ≥ n2).

∑ max{a2,a3,...,an}a1

≥ n.

Mihály Bencze

PP. 18750. Let a, b be squarefree positive integers and x, y, z ∈ N∗, suchthat x

√a+y

√b

y√a+z

√b∈ Q. Prove that 1xn+yn +

1yn+zn =

1yn for all n ∈ N

∗.

Mihály Bencze

PP. 18751. Compute1∫

−1arccos

(n∑k=1

(−1)k−1 x2k−1)dx.

Mihály Bencze

PP. 18752. Solve in R the equation

x11 + 3x10 + x9 + 3x8 + x7 − 3x6 − 17x5 + 3x4 + x3 + 3x2 + x+ 3 = 0.

Mihály Bencze

PP. 18753. Determinen∑k=1

ak if the sum of digits of the number

a1a2a3 + a2a3a4 + ...+ ana1a2 is odd.

Mihály Bencze

PP. 18754. Let be A =

1 7 x−x x 14 −1 −3

. Determine all x ∈ C such thatA2011 = 31670A.

Mihály Bencze

PP. 18755. Compute∫ (n+2)x2+(n+1)x+nx(x2+x+1)(xn+2+xn+1+xn+1)

dx on (0,+∞), wheren ∈ N.

Mihály Bencze


PP. 18756. Solve in [−2,+∞) the following system:(y − 1)2 (x+ 2)5 = 27 (z − 1)2 (2y + 1)

(y2 + y + 1

)(z − 1)2 (y + 2)5 = 27 (x− 1)2 (2z + 1)

(x2 + x+ 1

)(x− 1)2 (z + 2)5 = 27 (y − 1)2 (2x+ 1)

(y2 + y + 1

) .Mihály Bencze

PP. 18757. If 0 < a < b, then:

1). 54b∫a

x2+x+1(x+2)5

dx ≤ ln 2b+12a+1

2). 2916b∫a

x4+x2+1(x2+3x+2)5

dx ≤ ln (2b−1)(2a+1)(2b+1)(2a−1) .

Mihály Bencze

PP. 18758. Let be a, b ∈ R∗ and denote z ∈ C\R a root of the equationz3 + az + b = 0. Prove that a |z|6 + 8b |z|4Re(z) ≥ ab2.

Mihály Bencze

PP. 18759. If x ≥ −1, then(x2 + 3x+ 2

)5 ≥ 729 (4x2 − 1) (x4 + x2 + 1) .Mihály Bencze

PP. 18760. Solve in R the following system:(x+ 1)5 ≤ 27 (2y − 1)

(z2 − z + 1

)(y + 1)5 ≤ 27 (2z − 1)

(x2 − x+ 1

)(z + 1)5 ≤ 27 (2x− 1)

(y2 − y + 1

) .Mihály Bencze

PP. 18761. If xi > 0 (i = 1, 2, ..., k) ,Fn (x, y) = x

2n − x2n−1y + ...− xy2n−1 + y2n andSn =

(2n+1

1

)+ 2(2n+1

2

)+ ...+ n

(2n+1n

), then∑

cyclic

x1((x2x3)

SnF(2n+10 )n (x2,x3)F

(2n+11 )n−1 (x2,x3)...F

(2n+1n−1 )1 (x2,x3)

) 1n2n+1

≥ k.

Mihály Bencze


PP. 18762. Prove that(2n+1

0

)+(2n+1

1

)+(2n+1

2

)+ ...+

(2n+1n

)= 4n

2). Compute(kn+1

0

)+(kn+1

1

)+(kn+1

2

)+ ...+

(kn+1n

).

Mihály Bencze

PP. 18763. If x, y, z > 0, then∑ x

8√

(yz)3(y2−yz+z2)≥ 3.

Mihály Bencze

PP. 18764. If x, y, z > 0, then∑ x

44√

(yz)16(x2+z2)4(y4+z4)≥ 3

4√112 .

Mihály Bencze

PP. 18765. If x, y, z > 0, then∑ x64√

(yz)25(y2−yz+z2)5(y4−y2z+y2z2−yz3+z4)≥ 3.

Mihály Bencze

PP. 18766. If ak > 0 (k = 1, 2, ..., n) , An (a1, a2, ..., an) =1n

n∑k=1

ak,

Gn (a1, a2, ..., an) = n

√n∏k=1

ak, m = min {a1, a2, ..., an} ,

M = max {a1, a2, ..., an}, then determine all n ∈ N∗ for whichAn(a1,a2,...,an)Gn(a1,a2,...,an)

+ Gn(a1,a2,...,an)An(a1,a2,...,an) ≤mM +

Mn .

Mihály Bencze

PP. 18767. In all triangle ABC holds∑ (tgA2 +tgB2 )2(tg2 A2 +tg2 B2 )2

(tg2 A2 +6tgA2tgB

2+tg2 B

2 )2 ≥ 14 .

Mihály Bencze

PP. 18768. If ak > 0 (k = 1, 2, ..., n) andn∏k=1

ak = 1, then∏cyclic

a21+6a1a2+a22

(a1+a2)(a21+a22)≤ 2n.

Mihály Bencze


PP. 18769. Solve in R the following system:4x6 + 7x5 + 7y + 4 = 22z3

4y6 + 7y5 + 7z + 4 = 22x3

4z6 + 7z5 + 7x+ 4 = 22y3.

Mihály Bencze

PP. 18770. Let be a, k ∈ N∗, n ∈ N . Prove that:

1).(ka2 + 1

)2n+1can be expressed like a sum of k + 1 perfect squars

2).(ka2 + 1

)2n+2can be expressed like a sum of (k + 1)2 perfect squars

Mihály Bencze

PP. 18771. Solve in C the following system:(x21 + 6x1x2 + x

22

)√x2x3 = 2 (x1 + x3)

(x22 + x

23

)(x22 + 6x2x3 + x

23

)√x3x1 = 2 (x2 + x1)

(x23 + x

21

)(x23 + 6x3x1 + x

21

)√x1x2 = 2 (x3 + x2)

(x21 + x

22

)Mihály Bencze

PP. 18772. Prove that 102n+5 can be expressed like a sum of 25 perfectsquars, for all n ∈ N.

Mihály Bencze

PP. 18773. Prove that a, b, c are in arithmetical progression if and only ifa (2b) (2b) ... (2b)︸︷︷︸

n−time

c = 11...1︸︷︷︸(n+1)−time

ac.

Mihály Bencze

PP. 18774. Detemine all a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and all n ∈ Nfor which 2n + 1 is divisible by abc and 2n − 1 is divisible by abcd.

Mihály Bencze


PP. 18775. Determine the integer part of the expressionn∑k=1

√1 + 1

k2(k+1)2+ 1

(k2+k+1)2.

Mihály Bencze

PP. 18776. If xk > 0 (k = 1, 2, ..., n) , then

n∑k=1

√1 + x2k +

(xkxk+1

)2≥ n+

(n∑

k=1xk

)2n+

n∑k=1

xk

.

Mihály Bencze

PP. 18777. Let (akn)n≥1, (apn)n≥1 , (arn)n≥1 be arithmetical progressions.Determine all k, p, r ∈ C for which (an)n≥1 is an arithmetical progression.Some question for geometrical progression.

Mihály Bencze

PP. 18778. If a, b > 0 then solve in R the equation:√√√√√x+√x+ ...+

√x+

√x+ a︸︷︷︸

n−time

= b.

Mihály Bencze

PP. 18779. Let ABCD be a quadrilateral such that A = C = 90◦, andMk, Nk, Pk ∈ BD (k = 1, 2) where AM1, CM2 are altitudes; AN1, CN2 arebisectors; AP1, CP2 are medians. Prove thatAN1 + CN2 ≥ 12

(AM1 + CM2

)+ 2AD·AB

(AD+AB)2AP1 +

2CB·CD(CB+CD)2

CP2.

Mihály Bencze

PP. 18780. If a, b ∈ R, then determine all z ∈ C such that|z − a|z + 1|| = |z + b|z − 1||.

Mihály Bencze

PP. 18781. Prove that2

n(n+3)

n∑k=1

(k + 4)(k+2k+3

)3≥ 1 ≥ 2n(n+9)

n∑k=1

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

)2.

Mihály Bencze


PP. 18782. If a, b, c ∈ R such that∑a ∈ Q and (

∑a) (∑ab) = abc, then

3√∑

a3 ∈ Q.

Mihály Bencze

PP. 18783. If a, b, c > 0, then∑a3 + 12

∑ 1a3

≥√

6∑a3

abc .

Mihály Bencze

PP. 18784. If x ∈(0, π2

), then

14 +

(1+2 sin2 x)2

cos4 x+

(1+2 cos2 x)2

sin4 x≥ 24(4+sin

4 x+cos4 x)8+sin2 x cos2 x

.

Mihály Bencze

PP. 18785. In all triangle ABC holds:1). 5s

2+r2+4Rr4s2(s2+r2+2Rr)

≥ 1s2+r2+4Rr

+ 14(s2−r2−4Rr)

2). s2+r2+4Rr4s2Rr

≥ 1r(4R+r) +1

2(s2−2r2−8Rr)

3). (4R+r)2+s2

4s2R(4R+r)≥ 1

s2+ 1

2(4R+r)2−4s2

Mihály Bencze

PP. 18786. In all triangle ABC holds∑ √cos B

2+cos C

2−cos A

2√cos B

2+√

cos C2−√

cos A2

≤ 3.

Mihály Bencze

PP. 18787. If a, b, c > 0 and∑ab = 3, then

11+2abc

∑ 1a+1 ≥

abc1+abc

∑ 11+(b+c)a2

.

Mihály Bencze

PP. 18788. In all triangle ABC holds∑ sinA

2+cosA+cos(B−C) ≤ 1−sr8R2

.

Mihály Bencze

PP. 18789. If x ∈ R, then (1−√3 sinx)

2

1−9 sin4 x +(1−

√3 cosx)

2

1−9 cos4 x ≥7−2

√6

35 . When holdsthe equality?

Mihály Bencze


PP. 18790. If a, b, c > 0 then 3∑ a

b ≥ 7 +2∑a2∑ab .

Mihály Bencze

PP. 18791. If a, b, c > 0 and a2 + b2 + c2 = 1, then

∑√a2 +

(b2−c2

2

)2+ 2

∑a ≤ 3

√3.

Mihály Bencze

PP. 18792. Determine all x ∈ R and all n ∈ N such that(1n

n∑k=1

ak

)2+ xn

n∑k=1

ak ≥ n√

n∏k=1

ak

(1n

n∑k=1

√ak

)for all ak > 0

(k = 1, 2, ..., n) .

Mihály Bencze

PP. 18793. Let ABC be a triangle such that

a+A = b+B = c+ C = 2s+π3 . Prove that aB + bC + cA ≤(2s+π3

)2.

Mihály Bencze

PP. 18794. If xn+1 = xnxn−1...xn−k +k

√(xkn − 1)

(xkn−1 − 1

)...(xkn−k − 1

),

then determine all x1, x2, ..., xk ∈ R and k ∈ N for which k + k√k + kxn is a

perfect k-power.

Mihály Bencze

PP. 18795. Determine all x ∈ R such that 0 <√2011− ab <

xab for an

infinitely of pairs (a, b) of natural numbers.

Mihály Bencze

PP. 18796. Denote S (a) the sum of digits of a in decimal notation. Provethat there exist an infinity of n such that S (pnk) ≥ S

(pn+1k

)when pk denote

the k-th prime.

Mihály Bencze


PP. 18797. Prove that all the terms of the sequencea1 = a2 = ... = ak−1 = 1 andan+kan+1 = an+k−1an+2 + an+k−2an+3 + ...+ an+k−[ k2 ]+1

an+[ k2 ]are natural

numbers.

Mihály Bencze

PP. 18798. In all acute triangle ABC holds 24R5s ≤∑ 1

sinA+sinB sinC ≤6Rs .

Mihály Bencze

PP. 18799. Denote pn the n-th prime number. Prove that for an infinity of

natural numbers n the numerator of the fractionn∑k=1

1pk

has at least two

prime factors.

Mihály Bencze

PP. 18800. If Mk ={z ∈ C|zk = 1

}, then determine all ki, r ∈ N

(i = 1, 2, ..., n) such that Mk1 ∩Mk2 ∩ ... ∩Mkn =Mr.

Mihály Bencze

PP. 18801. 1). If xn =1

(n0)+ 1

(n1)+ ...+ 1

(nn), then

x3 + x4 + ...+ xn ≤ 2(n−2)(n−1)n + 4(13 +

14 + ...+

1n

).

2). Compute limn→∞

1n (x3 + x4 + ...+ xn) .

Mihály Bencze

PP. 18802. Solve on(0, π2

)the following system:

3√tgx+ 3

√tgy = 29 (10− sin 2z)

3√tgy + 3

√tgz = 29 (10− sin 2x)

3√tgz + 3

√tgx = 29 (10− sin 2y)

.

Mihály Bencze

PP. 18803. Solve in C the following system:|x− |y + 1|| = |z + |x− 1|||y − |z + 1|| = |x+ |y − 1|||z − |x+ 1|| = |y + |z − 1||

.

Mihály Bencze


PP. 18804. In all triangle ABC holds2∑

sin2A sin A2 cosB−C2 +

∑sin2A cosB cosC ≤

≤ (4R+r)2−s2(10R+3r)+4Rr(4R−r)

8R3.

Mihály Bencze

PP. 18805. In all triangle ABC holds3(4R+r)(s2+r2+4Rr)

4s2R≤ 5s2+r2+4Rr

s2+r2+2Rr.

Mihály Bencze

PP. 18806. Let be x, y ∈ R such that x2 + y2 = 1. Determinemax

n∑k=1

∣∣xk − yk∣∣ .Mihály Bencze

PP. 18807. In all triangle ABC holds∑ √a2+bc

b+c ≥ 36

√2Rr

s2+r2+2Rr.

Mihály Bencze

PP. 18808. Solve on(0, π2

)the following system:

3√tgx+ 3

√tgy =

√2(sin y+cos z)√

sin 2x

3√tgy + 3

√tgz =

√2(sin z+cosx)√

sin 2y

3√tgz + 3

√tgx =

√2(sinx+cos y)√

sin 2z

.

Mihály Bencze

PP. 18809. Let ABC be a triangle. Determine all λ ∈ R such that∑3

√ctgA2 + λctg

B2 ≤

rs .

Mihály Bencze

PP. 18810. In all triangle ABC holds∑ √cos2 A

2+cos2 B

2−cos2 C

2

cos A2+cos B

2−cos C

2

≤ 3.

Mihály Bencze

PP. 18811. Let be k ∈ N and k ≥ 2. Prove that are infinitely many positiveintegers n such that nk + 1 has a prime divisor greather than kn+ k

√kn.

Mihály Bencze


PP. 18812. If a, b > 0 and n ∈ N,n ≥ 2, then

1

n n√(π2 )

n−1

(1a +

1b

)<

n√

π2∫

0

(sinxn

a +cosxn

b

)dx < n

√π2 ·√

1a2

+ 1b2.

Mihály Bencze

PP. 18813. Prove that∞∑n=1

1(n+1)

√n> π√

6.

Mihály Bencze

PP. 18814. Prove thate−1−

n∑k=1

1k!(

n∑k=1

1k!

)n∏

k=1(k(e−1)−1)

≤ 1nn+1

.

Mihály Bencze

PP. 18815. If n > k ≥ 2, then there is a set of kn − 1 points in the plane,no (k + 1) collinear such that no kn form a convex kn−gon.

Mihály Bencze

PP. 18816. Let ABCD be a a tetrahedron inscribed in a sphere with centerO and radius R. AO meets the circumsphere of OBCD in A1, etc. Provethat OA1 ·OB1 ·OC1 ·OD1 ≥ 16R4

Mihály Bencze

PP. 18817. Let λ be the positive root of the equation x2 = 2011x+ 1, andbe m ∗ n = mn+ [λm] [λn] for all m,n ∈ N, when [·] denote the integer part.Prove that for all a, b, c ∈ N holds (a ∗ b) ∗ c = a ∗ (b ∗ c) .

Mihály Bencze

PP. 18818. Determine all integers n, k ∈ Z such that kn+1nk

∈ Z.

Mihály Bencze

PP. 18819. Find all ai, k ∈ N (i = 1, 2, ...., n) such thatk < a1 < a2 < ... < an and for which

n∏i=1

(ai − k) is a divisor ofn∏i=1

ai − kn.

Mihály Bencze


PP. 18820. Determine all k ∈ N∗ such that∞∑n=1

(√k +

3

√k + 1 + ...+ n

√n+ n+1

√n+ 1−

√k + 3

√k + 1 + ...+ n

√n

)<

< e− 1.

Mihály Bencze

PP. 18821. If ak > 0 (k = 1, 2, ..., ...) and ak + ak+2 > 2ak+1 (k = 1, 2, ...)

andk∑j=1

aj ≤ 1 for all k = 1, 2, ..., then∞∑k=1

(ak − ak+1) < π2

3 .

Mihály Bencze

PP. 18822. If ai ∈ Z (i = 1, 2, ..., kn) , (ai 6= aj if i 6= j) then determine all

x ∈ Z such thatkn∏i=1

(x− ai) = (−1)n (n!)k for all k ∈ N∗.

Mihály Bencze

PP. 18823. Determine all a, b, c, d > 0 such thatloga b < logb+1 (c+ 1) < logc+2 (d+ 2) < logd+3 (a+ 3) .

Mihály Bencze

PP. 18824. Solve the following system:x4 − y4 = 8

(y2 + yz + z2 + 1

)(z − x)

y4 − z4 = 8(z2 + zx+ x2 + 1

)(x− y)

z4 − x4 = 8(x2 + xy + y2 + 1

)(y − z)

.

Mihály Bencze

PP. 18825. Compute

{ ∑1≤i1


PP. 18827. If a0 =12 , ak+1 = ak +

a2kn (k = 0, 1, ..., n− 1), then[

m∑i=1

ai

]= m− 1 for all m ∈ N∗, where [·] denote the integer part.

Mihály Bencze

PP. 18828. Find all polynomials P for which

P (x)P(2x2)P(3x3)...P (nxn) = P

(n!x

n(n+1)2 + x+ 2x2 + ...+ nxn

).

Mihály Bencze

PP. 18829. Solve the following system:√2λ+ 1− x2 +

√3y + λ+ 4 =

√z2 + 9z + 3λ+ 9√

2λ+ 1− y2 +√3z + λ+ 4 =

√x2 + 9x+ 3λ+ 9√

2λ+ 1− z2 +√3x+ λ+ 4 =

√y2 + 9y + 3λ+ 9

, where λ ∈ R.

Mihály Bencze

PP. 18830. If xij > 0 (i = 1, 2, ..., n; j = 1, 2, ...,m) then determine all

c (n,m, k) > 0 such that 1 +n∑j=1

(m∑j=1

xij

)k≤ c (n,m, k)

m∏j=1

(1 +

n∑i=1

xkij

)for

all k ∈ N.

Mihály Bencze

PP. 18831. Prove that2n+1∑k=0

(2n+1k

) ((tg x2)2k

+(ctg x2

)2k+ (tgx)k

((tg x2)k

+(−ctg x2

)k))=

= 1(sin x2 )

4n+1 +1

(cos x2 )4n+1 .

Mihály Bencze

PP. 18832. Determine all positive integers such that its perfect k powers isequal to kn2 + kn+ 2k + 1, when n ∈ N.

Mihály Bencze

PP. 18833. Let ABC be a triangle. Determine all λ ∈ R for which ABC isrectangular triangle if and only if

(sinA)λ + (sinB)λ + (sinC)λ = λ((cosA)λ + (cosB)λ + (cosC)λ

).

Mihály Bencze


PP. 18834. Solve the following system:sin3 x+ sin3

(2π3 + y

)+ sin3

(4π3 + z

)+ 34 cos 2t =

= sin3 y + sin3(2π3 + z

)+ sin3

(4π3 + t

)+ 34 cos 2x = sin

3 z + sin3(2π3 + t

)+

+sin3(4π3 + x

)+ 34 cos 2y = sin

3 t+sin3(2π3 + x

)+sin3

(4π3 + y

)+ 34 cos 2z = 0.

Mihály Bencze

PP. 18835. Let ABC be a triangle such that

2s2

r

(s2 + r2 + 2Rr

)=∏

(atgA+ btgB). Prove that ABC is equilateral.

Mihály Bencze

PP. 18836. If x ∈(0, π2

], then

1 <(1− x22 +

x4

6

)2+(1− 12

(π2 − x

)2+ 116

(π2 − x

)4)2.

Mihály Bencze

PP. 18837. In all triangle ABC holds

2 + 2rR ≤∑(√

1 + sin 2A−√1− sin 2A

)≤ 3

√2.

Mihály Bencze

PP. 18838. Solve the following system√x2 + 2λx− λ2 = 1 +

√y2 − 2λy − λ2√

y2 + 2λy − λ2 = 1 +√z2 − 2λz − λ2√

z2 + 2λz − λ2 = 1 +√x2 − 2λx− λ2

, where λ ∈ R.

Mihály Bencze


4x2 < (2y + 9)

(1−

√2z + 1

)24y2 < (2z + 9)

(1−

√2x+ 1

)24z2 < (2x+ 9)

(1−

√2y + 1

)2 .Mihály Bencze


√3− x > 12 +

√y + 1√

3− y > 12 +√z + 1√

3− z > 12 +√x+ 1

Mihály Bencze


PP. 18841. Let beG =

{a+ b 5

√n+ c

5√n2 + d 5

√64 + e 5

√256|a, b, c, d, e ∈ Q

}\ {0} . Determine

all n ∈ N for which (G, ·) is abelian group.

Mihály Bencze

PP. 18842. Determine all m,n ∈ N for which (21n+4)(21m+25)(14n+3)(14m+17) isirreductibile.

Mihály Bencze

PP. 18843. If xk > 0 (k = 1, 2, ..., n) then

1).∑cyclic

(x21+x

22

x1+x2

)3≥

n∑k=1

x3k

2).∑cyclic

(x31+x32)(x1+x2)3

(x21+x22)2 ≤ 4

n∑k=1

x2k

Mihály Bencze

PP. 18844. If x, y, z ∈(0, π2

), then

4 + 2 (tgxtgytgz + ctgxctgyctgz) +√2(

1sinx sin y sin z +

1cosx cos y cos z

)≥

≥(

1sinx sin y sin z +

1cosx cos y cos z

)(sinx+ cosx) (sin y + cos y) (sin z + cos z) .

Mihály Bencze

PP. 18845. If xk > 0 (k = 1, 2, ..., n) then

max

{ ∑cyclic

(x1x2+x2x3+x3x1)3

(x1+x2)(x2+x3)2(x3+x1)

2 ;∑cyclic

(x1x2+x2x3+x3x1)3

(x1+x2)2(x2+x3)(x3+x1)

2 ;

∑cyclic

(x1x2+x2x3+x3x1)3

(x1+x2)2(x2+x3)

2(x3+x1)

}≤ 2732

n∑k=1

xk.

Mihály Bencze

PP. 18846. If xk > 0 (k = 1, 2, ..., n) then∑cyclic

11+8x1x2

< 18

n∑k=1

1x2k.

Mihály Bencze


PP. 18847. If xk > 0 (k = 1, 2, ..., n) thenn∑k=1

1xk

≥n∑k=1

11+xk

+ 4∑cyclic

11+8x1x2

.

Mihály Bencze

PP. 18848. If xk > 0 (k = 1, 2, ..., n) thenn∑k=1

1x2k

+ 2n∑k=1

1xk

≥ 2n∑k=1

11+xk

+ 16∑ 1

1+8x1x2.

Mihály Bencze

PP. 18849. Determine all n ∈ N and all prime p ≥ 3 for which both of thenumbers 2n − p and 2n + p are primes.

Mihály Bencze

PP. 18850. Determine all a, b, c ∈ R for which the equations

x2 + a2x+ b3 = 0, x2 + b2x+ c3 = 0, x2 + c2x+ a3 = 0 have a common realroot.

Mihály Bencze

PP. 18851. Let x = 0, a1a2a3... and y = 0, b1b2b3... be the decimalrepresentations of two positive real numbers. The equality bn = akn holds forall positive integers n. Determine all k ∈ N∗ for which x is rational number ifand only if y is rational number.

Mihály Bencze

PP. 18852. Determine all prime p for which 3p2 + 2, 4p3 + 1, 5p3 + 6p2 + 2are prime.

Mihály Bencze

PP. 18853. Let ABCD be a tetrahedron, and M ∈ Int (ABCD) . DenoteAM , BM , CM , DM the centers of circumspheres of tetrahedrons MBCD,MCDA, MDAB, MABC. Determine max V ol[AMBMCMDM ]V [ABCD] .

Mihály Bencze


PP. 18854. If n ∈ N, n ≥ 2, then

(n−1)2(n2+1)2(n+1)(((n2+1)(n−1))

n−1)n3−n2+n−2 +

n3(n4n−1)(n+1)(n2+1)

is divisible by n5 + 1.

Mihály Bencze

PP. 18855. Determine all ak ∈ Q∗ (k = 1, 2, ..., n) for whicha1 +

1a2a3...an

; a2 +1

a1a3...an; ...; an +

1a1a2...an−1

are integer numbers.

Mihály Bencze

PP. 18856. If a, b > 0, then

((1+a2)(1+b2)

a+b

)a+b≥(1 + a2

)b (1 + b2

)a.

Mihály Bencze

PP. 18857. Prove that any tetrahedron can be decomposed into ntetrahedrons which have three faces congruent isosceles triangles, for everyn ≥ 5.

Mihály Bencze

PP. 18858. Let ABCD be a trapezoid with AB paralell to CD andAB > CD. Let E and F be the points on the segments AB and CD,respectively, such AEEB = x

DFFC . Let K and L be two points on the segment

EF such that AKB] = yDCB] and CLD] = zCBA]. Determine allx, y, z ∈ R for which K,L,B,C are concyclic.

Mihály Bencze

PP. 18859. Let A1A2...An be a convex polygon inscribed in a circle withradius R. Denote B1, B2, ..., Bn the midpoints of sides A1A2, A2A3, ..., AnA1.Prove that

1).n∑k=1

1OBk

≥ nR cos πn

2).n∑k=1

OBk ≤ nR cos πn

Mihály Bencze


PP. 18860. If ak > 0 (k = 1, 2, ..., n) , then

n∏k=1

(1 + a2k

)≥ 2n−1n

∑cyclic

a1a2...an−1(1 + a2n

).

Mihály Bencze

PP. 18861. If an = k + a0a1...an−1 for all n ≥ 1, then determine alla0, k ∈ N for which (ai; aj) = 1, for all i, j ∈ N∗ (i 6= j) .

Mihály Bencze

PP. 18862. Let ABC be a triangle and M ∈ Int (ABC) ,CM ∩AB = {D} , D ∈ (AB) . The line segment through D perpendicularto BM intersect BC at E, the line segment through D parallel to BM meetsAC in F. Determine all points M for which E,M,F are collinear.

Mihály Bencze

PP. 18863. Solve in R the following equationn∑k=1

{kx} = [x] , where [·] and

{·} denote the integer, respective the fractional part.

Mihály Bencze

PP. 18864. Solve in R∗ the following equation

(n∑k=1

[kx

])( n∑k=1

[xk

])= n2,

where [·] denote the integer part.

Mihály Bencze

PP. 18865. Prove that 1n−

n∑k=1

cos(x+ kπn )+ 1

n+n∑

k=1cos(x+ kπn )

≤ 1sin2 nx

.

Mihály Bencze

PP. 18866. Solve in R the following equationn∏k=1

[xk]= x

n(n+1)2 , where [·]

denote the integer part.

Mihály Bencze


PP. 18867. If a, b > 0 and A = a+b2 ; G =√ab, K =

√a2+b2

2 , then

Gλ+Kλ

2 ≥ Aλ ≥

(G+K

2

)λfor all λ ∈ (−∞, 0] ∪ [2,+∞) .

Mihály Bencze

PP. 18868. Prove that5n∑k=1

1

sin2(

(12k+5)π60n

) = 5n2.Mihály Bencze

PP. 18869. If ak ≥ 1 (k = 1, 2, ..., n) , thenn∑k=1

11+ak

≤ n−12 +1

1+n∏

k=1ak

.

Mihály Bencze

PP. 18870. If a, b ∈ R, then∑cyclic

(|az1 + bz2|2 + ab |z1 − z2|2

)= (a+ b)2

n∑k=1

|zk|2 for all zk ∈ C

(k = 1, 2, ..., n) .

Mihály Bencze

PP. 18871. Solve the following equationn∑k=1

11+axk

= n−12 +1

1+n∏

k=1axk

, when

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

Mihály Bencze

PP. 18872. Let ABCD be a convex quadrilateral. Denote O1 and O2 thecircumcenter of triangles ABC and ADC, and A1, C1 the centers ofcircumcircles of triangles BO1C and AO1B, and A2, C2 the centers ofcircumcircles of triangles CO2D and AO2D. Compute

Area[A1A2C2C1]Area[ABCD] .

Mihály Bencze

PP. 18873. Let ABCD be a convex quadrilateral. Denote H1 theorthocenter of triangle ABC, and A1 is on the ray H1A such that AA1 = BCand B1, C1 is defined similar. Denote H2 the orthocentre of triangle ADC,and A2 is on the ray H2A such that AA2 = DC and C2, D2 is definedsimilar. Compute Area[A1A2B1C2C1D2]Area[ABCD] .

Mihály Bencze


PP. 18874. Let ABC be an acute triangle, with O the circumcenter.Denote A1, B1, C1 the circumcentres of triangles BOC,COA,AOB. ProveArea[A1B1C1]Area[ABC] =

R2

2(s2−(2R+r)2).

Mihály Bencze

PP. 18875. Let H be the orthocentre of the acute triangle ABC, with A1on the ray HA and such that AA1 = BC. Define B1, C1 similarly. Prove that2(2sr + 6Rr − 3r2

)≤ Area [A1B1C1] ≤ 2

(2R2 + 2sr + r2

).

Mihály Bencze

PP. 18876. If ak > 0 (k = 1, 2, ..., n) , then

n∑k=1

ank

n∏k=1

ak

+

n∑k=1

ak√n∑

k=1a2k

≥ n+√n.

Mihály Bencze

PP. 18877. If ak > 0 (k = 1, 2, ..., n) , then

2n∑k=1

a2k ≥∑cyclic

(a1a2

a1+a2+√

2(a21+a22)

)2.

Mihály Bencze


1). 2sinA + sinA2 + cos

A2 ≥ 2 +

√2 and its permutations

2). 3 + sR ≥∑(

2 +√2− 2sinA

)2Mihály Bencze

PP. 18879. If x > 0, then shx+ 1shx +1chx + thx ≥ 2 +

√2.

Mihály Bencze

PP. 18880. If x, y > 0 then(x2 + x+

(x2 + 1

)√x2 + 1

)(y2 + y +

(y2 + 1

)√y2 + 1

)≥

≥(6 + 4

√2)xy√

(x2 + 1) (y2 + 1).

Mihály Bencze



1). 3+4 sin2 A

2√3 sinA

+√3+2 sinA√3+4 sin2 A

≥ 2 +√2 and his permutations

2). 1+4 cos2 A

2 cosA +1+2 cosA√1+4 cos2 A

≥ 2 +√2 and his permutations

Mihály Bencze

PP. 18882. Solve in (0,+∞) the following systemx2 + x+

(y2 + 1

)√y2 + 1 =

(2 +

√2)z√z2 + 1

y2 + y +(z2 + 1

)√z2 + 1 =

(2 +

√2)x√x2 + 1

z2 + z +(x2 + 1

)√x2 + 1 =

(2 +

√2)y√y2 + 1

.

Mihály Bencze

PP. 18883. Let ABCD be a convex quadrilateral with a = AB, b = BC,c = CD, d = DA, e = BD, f = AC. Prove that 4

√3Area [ABCD] ≤

min{

(ab+bf+fa)2

a2+b2+f2+ (cd+df+fc)

2

c2+d2+f2; (ae+ed+da)

2

a2+e2+d2+ (bc+ce+eb)

2

b2+c2+e2

}.

Mihály Bencze

PP. 18884. If a0 6= 0 and a0zn + a1zn−1 + ...+ an−1z + an = 0 is apolynomial with complex coefficients and zeros z1, z2, ..., zn such that

|zk| < R (k = 1, 2, ..., n) , thenn∑k=1

∑1≤i1


PP. 18887. If ak > 0 (k = 1, 2, ..., n) andn∑k=1

1ak

= 1, then

∑cyclic

√a1a2...an−1 + an ≥ n

√nn−2+1

n+√nn−1

(n

√n∏k=1

ak +n∑k=1

√ak

).

Mihály Bencze

PP. 18888. If ak > 0 (k = 1, 2, ..., n) then

nn∑k=1

ak ≥(n2 − 1

)n

√n∏k=1

ak + n

√1n

n∑k=1

ank .

Mihály Bencze

PP. 18889. Let be A = aa...a︸︷︷︸n−time

bb...b︸︷︷︸m−time

. Determine all

a, b ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and all k ∈ N for which Ak is palindrome.

Mihály Bencze

PP. 18890. Let ABC be an acute triangle and M ∈ Int (ABC) . For eachof triangles ABM,BCM,CAM draw its Euler line, that is, the lineconnecting its circumcentre and its centroid. Determine all points M forthese three lines are concurent.

Mihály Bencze

PP. 18891. If ak > 0 (k = 1, 2, ..., n) then

n∑k=1

an−1k +n2(n−2)

n∏k=1

ak

n∑k=1

ak

≥ (n− 1)∑cyclic

a1a2...an−1.

Mihály Bencze

PP. 18892. Solve in N the following system:

σ (d (n)) = mσ (d (m)) = kσ (d (k)) = n

.

Mihály Bencze

PP. 18893. In all triangle ABC holds∑( 1a2(s−a)2+s2r2

)λ≥ 31−λ

(4R+r

s2r2(4R−r)

)λfor all λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihály Bencze


PP. 18894. In all triangle ABC holds3

√(4R+r)3−12s2R

3s2r+ s√

(4R+r)2−2s2≥ 2.

Mihály Bencze

PP. 18895. If xk > 0 (k = 1, 2, ..., n) , then

n

√√√√√ n∑k=1xnkn

n∏k=1

xk

+ n−1

√√√√ ∑cyclicx1x2...xn−1n∑k=1

xn−1k

≥ 2.

Mihály Bencze

PP. 18896. If xk > 0 (k = 1, 2, ..., n) such thatn∏k=1

xk = 1 then

(n− 1)√n (xx21 + x

x32 + ...+ x

x1n ) +

n∑k=1

xk ≥ n2.

Mihály Bencze

PP. 18897. Compute the following sum:n∑k=1

[3k2+2k

√k+

3√k

k+6

], when [·]

denote the integer part.

Mihály Bencze

PP. 18898. If x ∈(0, π2

), then sin

8 x+ctg8xcos6 x

+ cos8 x+tg8xsin6 x

≥ 8.

Mihály Bencze

PP. 18899. Prove that∞∑k=1

(e√k4+1−k2 − 1

)> π

2

12 .

Mihály Bencze

PP. 18900. In all triangle ABC holds 5∑tg2A2 tg

2B2 ≤ 1 + 6

∑tg3A2 tg

3B2 .

Mihály Bencze

PP. 18901. In all triangle ABC holds∑ sinA sin2B

(4 sin2B−sinA sinC) sinC≥ 1.

Mihály Bencze



sin A2

≤ 2(2R−r)3r .

Mihály Bencze


(1− sinA) (1− sinB) (1− sinC) ≤ (26−15√3)r2

2R2,

Mihály Bencze

PP. 18904. If x > 0 then in all triangle ABC holds∏(sin A2 + cos

A2

) (1− sin A2 cos

A2

)≥ (1+x

2)(s+xr)4R(1+x3)

.

Mihály Bencze

PP. 18905. In all triangle ABC holds∏(

tg2A2 + tg2B2

)≥ 8r

2((4R+r)2−2s2)2

s6.

Mihály Bencze

PP. 18906. In all triangle ABC holds∑ctgA2

√1 +

(tgB2 − tg

A2

)tgC2 ≤

sr .

Mihály Bencze


3−tgB2tgC

2+tg2 A

2

≤ 1.

Mihály Bencze

PP. 18908. In all triangle ABC holds∑√

tgA2(1− tgA2 tg

C2

)≤√

2s3r .

Mihály Bencze


a1√a2+a3+...+an

≥√

nn−1

n∑k=1

ak.

Mihály Bencze

PP. 18910. In all triangle ABC holds s(R+2r)R2

+∑

sin 3A < 9√3

2 .

Mihály Bencze


PP. 18911. If x ∈[0, π2

]then 2 cosx

√sinx+

√sinx(1 + 4 sin2 x) ≤ 3

4√3√2.

Mihály Bencze

PP. 18912. If x ∈(0, π2

), then

(1 + sinx) (1 + cosx)(

1sinx +

1cosx

)≥ 4 + 3

√2.

Mihály Bencze

PP. 18913. In all triangle ABC holds∑ √ctgA

2

sn−(rctgA2 )n ≥ (2n+1)

√s 2n

√2n+1

2nsn√r

.

Mihály Bencze

PP. 18914. In all triangle ABC holds1∑sin2 A

+ 1∑cos2 A

≥ 23 +∏

sin2A+∏

cos2A.

Mihály Bencze

PP. 18915. In all triangle ABC holds∑(

3 + 16 cos4A)4

+ 256 ≥ 2048rR .

Mihály Bencze

PP. 18916. If ak > 0 (k = 1, 2, ..., n) , then(n∑k=1

ank

)(n∑k=1

1ank

)≥ nn−1

∑cyclic

a2+a3+...+ana1

.

Mihály Bencze

PP. 18917. In all triangle ABC holds∑(

ctgA2)nctgB2 ≤

nnsn+1

rn+1(n+1)n+1for

all n ≥ 2.

Mihály Bencze

PP. 18918. In all triangle ABC holds∑√36r2ctg2A2 +

s2tgA2

6(tgB2 +tgC2 )

≥ 3s√17

2 .

Mihály Bencze

PP. 18919. In all triangle ABC holds 9 (2R+ r) ≥ 5s√3.

Mihály Bencze


PP. 18920. In all triangle ABC holds∑( cos A

2

cos B2

)2≤ 72 .

Mihály Bencze

PP. 18921. If x ∈(0, π2

), then 1

sin6 x+cos6 x+ 4

sin2 2x≥ 4 + 2

√3.

Mihály Bencze

PP. 18922. In all triangle ABC holds∏(tg2A2 + x

(tg2B2 + tg

2C2

))≥ (1+2x)

√1+8x−1−6x4 for all x ≥ −

18 .

Mihály Bencze

PP. 18923. If ak > 1 (k = 1, 2, ..., n) , then

aloga2 a31 + a

loga3 a42 + ...+ a

loga1 a2n ≥ n n

√n∏k=1

ak.

Mihály Bencze

PP. 18924. In all triangle ABC holds∑ (ctgA2 )6

(ctgB2 )3+(ctgC2 )

3 ≥ s3

18r3.

Mihály Bencze

PP. 18925. In all triangle ABC holds∑√ tgA

2

tgA2+ctgA

2

≤ 32 .

Mihály Bencze

PP. 18926. If x ∈ R then 2 cos2 x+ 34 sin2 x

+ 54 cos2 x

≥ 5.

Mihály Bencze

PP. 18927. In all triangle ABC holds∑

5

√2ctgA2 + ctg

B2 ≤ 3 5

√sr .

Mihály Bencze

PP. 18928. If r ∈ (0, 1] then 1sin6 x+cos6 x

+ 1sin2 x cosx

+ 1sinx cos2 x

≥ 20r3 forall x ∈

(0, π2

).

Mihály Bencze


PP. 18929. In all triangle ABC holds∑5

√(2ctgA2 + ctg

B2

) (ctgA2 + ctg

C2

)ctgA2 ≤

5

√54(sr

)3.

Mihály Bencze

PP. 18930. In all triangle ABC holds∑ tg8 A

2

(tg2 A2 +tg2B2 )

2 ≥ 112 .

Mihály Bencze

PP. 18931. In all triangle ABC holds∑ ctgA

2

s3+sr2ctg2 A2+r3ctg3 A

2

≤ 2713s2r

.

Mihály Bencze


tg2A2 + tgA2 tg

B2 + tg

2B2 ≥

3√3s

4R+r .

Mihály Bencze

PP. 18933. In all triangle ABC holds 512s2Rr ≤(s2 + r2 + 2Rr

)2.

Mihály Bencze


s−rctgA2

≥ 2∑ 1

s+rctgA2

.

Mihály Bencze

PP. 18935. In all triangle ABC holds∑ ctgA

2

rctg2 B2+sctgC

2

≥ 94s .

Mihály Bencze


s+4rtgA2

≥ 2713s .

Mihály Bencze

PP. 18937. In all triangle ABC holds∑tg2A2 tg

2B2 ≤

12 −

(2rs

)2.

Mihály Bencze

PP. 18938. If ak ≥ 1 (k = 1, 2, ..., n) , then∑cyclic

(3(a21−a22

8

)2+ a1a2a1+a2

)2≥ 14

n∑k=1

a2k.

Mihály Bencze


PP. 18939. In all triangle ABC holds∑√ctgA2 − ctg

B2 − ctg

C2 +

rs ≥ 6

√rs .

Mihály Bencze

PP. 18940. In all triangle ABC holds∑(tg2A+ ctg2A

)≥ 3

3√

(s2−(2R+r))2

3√

(4R2)2− 3√(s2−(2R+r)2)

2+

3 3√

(sr)2

3√

(2R2)2− 3√

(sr)2

Mihály Bencze

PP. 18941. If a, b > 0 and x ∈ R, then(a sin2 x+ b cos2 x

) (sin2 xa +

cos2 xb

)≥ 1.

Mihály Bencze


tgA2 tgB2 ≤

s3r .

Mihály Bencze

PP. 18943. In all triangle ABC holds s (∑√

a) (∑a√a) ≥ 2

∑a3.

Mihály Bencze

PP. 18944. In all triangle ABC holds(s2+r2+4Rr

4Rr

)2≥ 1Rr + 4

(∑ tgA2tgB

21+a2

)2.

Mihály Bencze

PP. 18945. Compute limn→∞

1n

n∑k=2

logk (k + 1) .

Mihály Bencze

PP. 18946. Determine all x, y > 0 such that 1n∑k=1

1x+ak

− 1n∑k=1

1y+ak

≥ 1n .

Mihály Bencze

PP. 18947. In all triangle ABC holds 1 +∑tg2A2 tg

2B2 ≥

4r(4R+r)s2

.

Mihály Bencze


PP. 18948. In all triangle ABC holds∑ A sinB

sinA ≤ π.

Mihály Bencze

PP. 18949. In all triangle ABC holds∑ A cos B−C

2

sin A2

≤ π.

Mihály Bencze

PP. 18950. Prove that√2 3√

3 4√

4... n√n+

√1 +

√3 +

√5 + ...+

√2n− 1 < 4.

Mihály Bencze

PP. 18951. Prove that∞∑n=1

(4n−1∑k=1

1(2k−1)(4n−k)

)2< π

2

6 .

Mihály Bencze

PP. 18952. Prove thatm∏p=3

p∏k=2

(1− 1kp

)> 3m+1 .

Mihály Bencze

PP. 18953. In all triangle ABC holds∑ tgA

2√tg2 A

2+ 2r

sctgA

2

≤ 4R+rs .

Mihály Bencze

PP. 18954. In all triangle ABC holds1). 5s

2+r2+4Rr4s2(s2+r2+2Rr)

≥ 1s2+r2+4Rr

+ 14(s2−r2−4Rr)

2). s2+r2+4Rr4s2Rr

≥ 1r(4R+r) +1

2(s2−2r2−8Rr)

Mihály Bencze

PP. 18955. In all triangle ABC holds∑ s2+r2ctg2 A

2

(tgA2 +tgB2 )

2 ≥ 5s2

2 .

Mihály Bencze

PP. 18956. If a, b, c > 0, then determine all x, y > 0 such that∑(a+xba+yc

)3≥ 3.

Mihály Bencze


PP. 18957. If x, y, z ∈(0, π2

), then

(∑

sinx cos y cos z) (∑

cosx sin y sin z) ≤ (1 +∏

sinx) (1 +∏

cosx) .

Mihály Bencze

PP. 18958. In all triangle ABC holds∑ (s+3rctgA2 )ctg2 C2

ctgA2+ctgC

2+3tgB

2

≥ 2s23r .

Mihály Bencze


2

3r2ctg2 B2+2rsctgC

2+3s2

≤ 112sr .

Mihály Bencze

PP. 18960. In all triangle ABC holds∑ 3√tg2 A

2tg2 B

2√1− 3√tg2 A

2tg2 B

2

≥ 2.

Mihály Bencze


2tgB

2

1+√tgB

2tgC

2−√tgA

2tgB

2

≥ 1.

Mihály Bencze


cos A2

≥ 2∑√

tgA2 tgB2 .

Mihály Bencze

PP. 18963. If ak > 0 (k = 1, 2, ..., n) , then∏cyclic

(a1+a2a3...an

1+a1

)≥

n∏k=1

ak.

Mihály Bencze

PP. 18964. In all triangle ABC holds∑ 1√

s+rtgA2

≥ 9√10s.

Mihály Bencze

PP. 18965. Determine all f : R→ R for whichx7y7 (f (x+ y)− f (x)− f (y)) == 5f (x) f (y)

(x2f (y) + y2f (x) + 2

(x4f (y) + y4f (x)

))for all x, y ∈ R.

Mihály Bencze


PP. 18966. If a, b ∈ R∗ and z is a strictly complex root of the equationxn + ax+ (n− 1) b = 0, then |z| ≥ n

√|b|.

Mihály Bencze

PP. 18967. If p > 2 is a prime and z = cos 2πp + i sin2πp such that

11−z = a0 + a1z + a2z

2 + ...+ ap−2zp−2, then

1). a0 + a1 + ...+ ap−2 =p−12

2). 1 + pn−1a0a1...ap−2 is divisible by p

Mihály Bencze

PP. 18968. Solve in R the following system:sinx+ sin y + sin z = 0cosx+ cos y + cos z = 0

tg3kx+ tg3ky + tg3kz = 2(2−

√3) , where k ∈ N.

Mihály Bencze

PP. 18969. Let bexn − 1λx

n−1 + a1xn−2 − a2xn−3 + ...+ (−1)n−1 an−2x+ (−1)n an−1 = 0, where

ak ∈ C (k = 1, 2, ..., n− 1) and∣∣a1 + λa2 + λ2a3 + ...+ λn−2an−1∣∣ ≤ 1λ2 ,

where λ > 0. Prove that exist at least one root x1 of the given equation forwhich |x1| ≤ 2λ .

Mihály Bencze

PP. 18970. The numbers z1, z2 ∈ C∗ have the property (M) if existλ ∈ [−2, 2] such that z21 + z22 = λz1z2. Prove that z

n−k1 z

k2 , z

k1z

n−k2

(k = 0, 1, ..., n) have the property (M) too.

Mihály Bencze

PP. 18971. Let be az2 + bz + c = 0, where a, b, c ∈ C∗ and z1, z2 denote theroots of the given equation. If z1 · z2 = λ2, when λ > 0 then λ |b|+ |c| ≥ 3λ2.

Mihály Bencze

PP. 18972. Solve in C the equation2n+1∑k=0

|z − k| = (n− 1)2 .

Mihály Bencze


PP. 18973. We consider the equation az2 + bz + c = 0 when a, b, c ∈ C∗ and|a| = |b| = |c| . If the equation have a root with modul r > 0 and2t ∈

{−r4 + r2 − 1±

√(r4 + r2 + 1) (r4 − 3r2 + 1)

}, then b2t = (t+ 1)2 ac.

Mihály Bencze

PP. 18974. Solve in R the equation 2x + 5x + 20x + 50x = 4582175 · 10x.

Mihály Bencze

PP. 18975. Solve in R the equation8x + 27x + 125x + 343x = 30x + 42x + 720x + 105x.

Mihály Bencze

PP. 18976. If 1 < a ≤ 10 then solve in R the equationalg x + 100 = a2 +

(a2 + x− 100

) 1lg a .

Mihály Bencze

PP. 18977. Solve in R the equation 3x + 2 · 5x + 7x = 92105 |x|3 .

Mihály Bencze

PP. 18978. Solve in R the equation 32x+1 − (x− 3) 3x = 10x2 + 13x+ 4.

Mihály Bencze

PP. 18979. Solve in R the following equation(3lg 10x + 5lg 10x

)2= 34

(9lg x + 25lg x

).

Mihály Bencze

PP. 18980. Solve in[−13 ,+∞

)the following system:{

43x+1 + 43y+1 + 43z+1 + 43t+1 = 1024(3x+ 1) (3y + 1) (3z + 1) (3t+ 1) = 256

.

Mihály Bencze

PP. 18981. Solve in R the following equationx (x+ 2) (x+ 8) (x+ 26) (x+ 80) = 3y.

Mihály Bencze


PP. 18982. Solve in (0,+∞) the following system:ex

λ+ (λ− 1) lnx = ey

eyλ+ (λ− 1) ln y = ez

ezλ+ (λ− 1) ln z = ex

, when λ ≥ 1.

Mihály Bencze

PP. 18983. Solve in R the following equation lg(12x

3 + 13x2 + 16x

)= log2 x.

Mihály Bencze

PP. 18984. Solve in R the following equation3x

3+11x + x6 + 85x2 = 36x2+3 + 14x4 + 9.

Mihály Bencze

PP. 18985. If n ∈ N,n ≥ 2 then solve the following equation:1x + 2x + ...+ (n− 1)x = (n+ 1)x + (n+ 2)x + ...+ (2n− 1)x .

Mihály Bencze

PP. 18986. Solve in R the following equation 3√x2+5

√x−3 = 18 + 6x− x2.

Mihály Bencze

PP. 18987. Solve in [0,+∞) the following system:x+

√1 + nx+ (n− 1)x2 + ...+ 2xn−1 + xn = 10y−x

y +√

1 + ny + (n− 1) y2 + ...+ 2yn−1 + yn = 10z−yz +

√1 + nz + (n− 1) z2 + ...+ 2zn−1 + zn = 10x−z

when n ∈ N∗.

Mihály Bencze

PP. 18988. Computen∑k=0

(n+m+p

k

)2n−k +

m∑k=0

(n+m+p

k

)2m−k +

p∑k=0

(n+m+p

k

)2p−k.

Mihály Bencze

PP. 18989. Determine all n, p ∈ N∗ for which

n∑k=0

k(k+1)2(nk)n∑

k=0k(k+p)(nk)

∈ N.

Mihály Bencze


PP. 18990. Determine all n ∈ N∗ for which

n∑k=0

k2

(k+1)(k+2)(nk)

n∑k=0

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

nk)

∈ N.

Mihály Bencze

PP. 18991. Solve in N the following equationn∑k=0

(k + 1)2 (k + 2)2(nk

)= 2m.

Mihály Bencze

PP. 18992. Prove that[n2 ]∑k=0

∣∣∣(nk)− ( nk+1)∣∣∣ ≤√

[n2 ]+1n+1

(2nn

).

Mihály Bencze

PP. 18993. Prove thatm∏k=0

((mk

)(nk

))2k ≤ ( 78m+1−1

m∑k=0

(mk

)(n+kn

)) 8m+1−17.

Mihály Bencze

PP. 18994. Prove that

(4n+1

(n+1)(2n+1−1)

n∑k=0

(nk

)−1)1−( 12)n+1 ≥ n+1∏k=1

(1k

)2−k.

Mihály Bencze

PP. 18995. If n,m ∈ N∗, then

(2n+1 − 1

)( n∏k=0

(m+kk

)2n−k) 12n+1−1+(2m+1 − 1

)( m∏k=0

(n+kk

)2m−k) 12m+1−1 ≤≤ 2n+m+1.

Mihály Bencze

PP. 18996. If xn =1

k−1 −n∑

m=1

1

(k+mk ), where k ∈ N, k ≥ 2, then compute

limn→∞

nk−1xn and limn→∞

n(nk−1xn − k!k−1

).

Mihály Bencze


PP. 18997. If Sn =1

n+1

n∑k=0

1

(nk), then

1).n∏k=2

Sk ≥ 2(n+2)!S2n+1

2). 2n−1√Sn ≥ 1 +

n−2∑k=1

2k√

2k+2 .

Mihály Bencze

PP. 18998. In all acute triangle ABC holds(∑ 1

a

) (∑ a+bab−1

)≥ 9

José Luis Dı́az-Barrero

PP. 18999. Compute limn→∞

n∏i=1

(i∑

k=1

3k2+9k+7(k+1)3(k+2)3

).


PP. 19000. Find the values of a for which the function

f (z) =x∫1

(2z2

+ az4

)sin zdz is singled-valued.


PP. 19001. If 0 < a < b and A (x) is a polynomial with real coefficients forwhich A (a) <

√b− a < A (b) , then show that there exist two real numbers

λ1, λ2 ∈ (a, b) such thatb∫aA (x) dx = A (λ1)A

2 (λ2) .


PP. 19002. Solve in R the following system:

5a2 = c6 + 25b2 = a6 + 25c2 = b6 + 2

.


PP. 19003. Let the eulerian integrals I1 =1∫

−∞

dxxα and I2 =

5−0∫3

dx(5−x)β

. I1

and I2 are simultan convergents if:1). α = β = 12). α > 1, β > 1,3). α > 1, β < 1


4). α < 1, β > 1

5). α > 0, β > 0

Laurenţiu Modan

PP. 19004. Solve the following equation 1 + x√

x−1x+1 = (x+ 1)

√xx+2 .

György Szöllősy

PP. 19005. Let the function sequence (fn (x))n≥0 be, where fn : R→ R forall n ∈ N, and fn (x) = x1+(nx)2 .Study the simple and the uniform convergence for (fn (x))n≥1 .

Study the simple and the uniform convergence for the funcdtion series∑n≥0

x1+(nx)2

.

Laurenţiu Modan

PP. 19006. Find∞∑n=1

xn2

n for all x ≥ 0.

Laurenţiu Modan

PP. 19007. If the points A1, B1, C1 divides the sides BC,CA,AB oftriangle ABC in the same ratio k > 0, then

∑AA21 ≥ 34

∑a2.

Florentin Smarandache

PP. 19008. In a triangle ABC we draw the Cevians AA1, BB1, CC1 thatintersect in P .

Prove that∏ PA

PA1=∏ AB

A1B.


PP. 19009. In triangle ABC let‘s consider the Cevians AA1, BB1, CC1 thatintersect in P (A1 ∈ BC;B1 ∈ CA;C1 ∈ AB) .Prove that

1).∑ PA

PA1≥ 6

2).∏ PA

PA1≥ 8



PP. 19010. Let‘s consider a polygon (which has at least 4 sides)circumscribed to a circle, and D the set of its diagonals and the lines joiningthe points of contact two non-adjacent sides. Then D contain at least 3concurent lines.


PP. 19011. In all triangle ABC holds∑ ma(1+cosA cos(B−C)) sinA ≤ R

∑ 1sinA cos B−C

2

.


PP. 19012. If a, b > 0 and x ∈ [0, 1], then√

aa+xb +

√b

b+xa ≤2√1+x

.


PP. 19013. Solve the equation (1 + lnx)2 = 49√ex2.


PP. 19014. Solve the equation 2x4 − 5x2 + 1 + 14(sin πx2

)2= 0.


PP. 19015. If ak > 0 (k = 1, 2, ..., n) , then∑ a1√

a2+a3+...+an≥√

nn−1

n∑k=1

ak


PP. 19016. In all triangle ABC holds∑ a

b +√

1 + 3abc∑a2b

≥ 3 +√2.

D.M. Milošević

PP. 19017. If A = a+b2 , G =√ab, H = 21

a+ 1

b

, then H2 ≥ A (4G− 3A) .

D.M. Milošević

PP. 19018. If A = a+b2 , G =√ab, H = 21

a+ 1

b

, K =√

a2+b2

2 , then

1). A2 ≥ KG2). A+H ≥ 2G


3). K +G ≤ 2A4). 2K +H ≤ 3A

D.M. Milošević

PP. 19019. In all triangle ABC holds∑ A

B ≥(a+b+c3√9abc

)3.

D.M. Milošević

PP. 19020. Let ABC be a triangle such that C = 90◦. Prove that:1). a

2

m2a+ b

2

m2b≥ 53

2). a2

w2a+ b

2

w2b≥ 32 +

r2R

D.M. Milošević

PP. 19021. In all triangle ABC holds∑ rarb

c2≥(2− rR

)2.

D.M. Milošević

PP. 19022. Let x, y, z be the distances from the centroid of triangle ABC tothe sides BC,CA,AB respectively. Show that

√x+

√y +

√z ≤

√3 (R+ r).

D.M. Milošević

PP. 19023. In right-angled triangle ABC holds:1). 32 <

a2

w2b+ b

2

w2a≤ 12

(2 +

√2)

2). awb +bwa

≤√

2 +√2

D.M. Milošević

PP. 19024. If 0 < x < π2 , thensin2 x1+sinx +

cos2 x1+cosx ≤ 2−

√2.

D.M. Milošević

PP. 19025. If A = a+b+c3 , G =3√abc, H = 31

a+ 1

b+ 1

c

, K =√

a2+b2+c2

3 , then

K2H ≥ A(3AH − 2G2

).

D.M. Milošević

PP. 19026. If n,m ∈ N, 0 ≤ n ≤ 4, then m+ 1m +√

2mnm2+1

≥ 2 +√n.

D.M. Milošević


PP. 19027. In every triangle the following equalities are valid:

1).∑ ra+rb

rc= 2(2R−r)r

2).∑ ra+rb

a+b =2s(3R+2r)s2+2Rr+r2

D.M. Milošević


1).∑ ha

ra−r ≥ 2(2− rR

)22).

∑w2a ≥ 12

(s2 + 12Rr + 3r2

)D.M. Milošević

PP. 19029. If a, b > 0, n ∈ N,n ≥ 2, then1

n

√n∑

k=0(nk)

2(ab )

k+ 1

n

√n∑

k=0(nk)

2( ba)

k= 1

n

√√√√ [n2 ]∑k=0

( n2k)2(

ab

(a+b)2

)k . Compute

[n2 ]∑k=0

14k(k!)2(n−2k)! .


PP. 19030. If x, y, z ∈ R∗, x+ y ≥ 0, y + z ≥ 0, z + x ≥ 0, then1).

∑ x3x2+xy+y2

≥ 13∑x

2).∑ x3+y3

x2+xy+y2≥ 23

∑x

Mihály Bencze and Ovidiu Pop

PP. 19031. If n ∈ N,n ≥ 2, then xn+1xn+yn ≥2x−y2 for x, y ∈ R, y ≥ 0,

xn + yn 6= 0 if and only if n = 2.

Mihály Bencze and Ovidiu Pop

PP. 19032. The incircle of triangle ABC touches the lines BC, CA andAB at D,E and F respectivelly.Prove that

1). If a 6= b 6= c, then∑ AD2−BE2

b−a =2s(R+r)

R2). If a 6= b 6= c, then∑(

bn−1 + bn−2a+ ...+ ban−2 + an−1)ab(AD2 −BE2

)= 0 for all n ∈ N∗

3). The triangle ABC is isosceles if and only if two of AD,BE,CF are equal

4). The triangle ABC is equilateral if and only if AD = BE = CF

Mihály Bencze


PP. 19033. If a, b, c > 0, then∑√ 2

a +1b +

cab ≥

a+b+c√abc

+∑ 1√

a.

Mihály Bencze

PP. 19034. Let be xn (xn−1 + xn+1) = 2xn−1xn+1 for all n ≥ 1. Prove thatif some terms xi of the given sequence are the k power of a rational number,then the given sequence contains an infinite number of terms xj that areeach k powers of rational numbers.

Mihály Bencze

PP. 19035. Let be 1 = d1 < d2 < ... < dk = n the divisors of n. Determineall n ∈ N∗ and all i ∈ {1, 2, ..., k − 1} such that d2i + d2k−i = 2n+ 1.

Mihály Bencze

PP. 19036. If x, y, z > 0, then 3√∑

x3

3xyz +3

√∑x3y3

3x2y2z2+√∑

xy∑x2

+√

xyz∑x∑

x2y2≥ 4.

Mihály Bencze

PP. 19037. Determine all functions f : R→ R such that(f (x+ y)− f (x)− f (y))x3y3 = 4

(x2 + y2

)f (xy) + 6xyf (x) f (y) for all

x, y ∈ R.

Mihály Bencze

PP. 19038. Determine all prime p for which 5p2 + 2 and 3p3 + 2 are primetoo.

Mihály Bencze

PP. 19039. Solve in Z the system:

{xy + yz + zx ≡ 1 (mod n)x2 + y2 + z2 ≡ 1 (mod n) , where

n ∈ N,n ≥ 2.

Mihály Bencze

PP. 19040. If a, b, c > 0, then

a2 + b2 + c2 ≥ ab+ bc+ ca+ 14

((a2−b2)

2

a2+b2+

(b2−c2)2

b2+c2+

(c2−a2)2

c2+a2

). (A

refinement of the inequality∑a2 ≥

∑ab).

Mihály Bencze


PP. 19041. If ak > 0 (k = 1, 2, ..., n) , thenn∏k=1

(1− ak + a2k

)≥

(12

n

√n∏k=1

(1 + a4k

)+ 12

n

√n∏k=1

(1− a4k

))n.

Mihály Bencze

PP. 19042. If ak > 0 (k = 1, 2, ..., n) , then 2n

n∏k=1

(1 + a2k

)3 ≥≥

(n

√n∏k=1

(1 + ak)3 (1 + a3k)+ n

√n∏k=1

(1− ak)4(1 + ak + a

2k

))n.

Mihály Bencze

PP. 19043. If x ∈ R, then 11+3 sin4 x

+ 11+3 cos4 x

≥ 87 .

Mihály Bencze

PP. 19044. Solve in R the following system:x61 + 11x

22

(x22 + 1

)+ 1 = x3

(5x43 + 11x

23 + 5

)x62 + 11x

23

(x23 + 1

)+ 1 = x4

(5x44 + 11x

24 + 5

)−−−−−−−−−−−−−−−−−−−−x6n + 11x

21

(x21 + 1

)+ 1 = x2

(5x42 + 11x

23 + 5

) .Mihály Bencze

PP. 19045. Solve in (0,+∞) the following system:2+x21

(1+x1)2 +

x2(1+x2)

2 =2+x22

(1+x2)2 +

x3(1+x3)

2 = ... =2+x2n

(1+xn)2 +

x1(1+x1)

2 =34 .

Mihály Bencze

PP. 19046. If x ∈ R then 25−4 sin2 x cos2 x +

14+sin2 x

+ 14+cos2 x

≤ 1.

Mihály Bencze

PP. 19047. If x ∈ R then 12(1−sin2 x cos2 x)

+ 13+sin2 x

+ 13+cos2 x

≤ 1511 .

Mihály Bencze

PP. 19048. If x ∈(0, π2

), then 5 sin2 x+ 32

sin2 x≥ 37.

Mihály Bencze


PP. 19049. If x ∈(0, π2

), then

19−sin 2x +

17−2

√2 sinx+cos2 x

+ 17−2

√2 cosx+sin2 x

≤ 12 .

Mihály Bencze

PP. 19050. If x ∈(0, π2

), then

(√2 sinx cosx

)√2 (5− 2 sin2 x cos2 x

)≤ 3.

Mihály Bencze

PP. 19051. If x > 0 and t ≥ 3 then xt + x−t ≥ 2t+ (t+ 1)(x+ 1x

).

Mihály Bencze

PP. 19052. If x > 0 then(23

)x+(23

) 1x ≤ 43 .

Mihály Bencze

PP. 19053. If x ∈ R, then 2(sin6 x+ sin6 x

)≤ 3 + 7

(sin4 x+ cos4 x

).

Mihály Bencze

PP. 19054. If x, y, z ∈ R then2 + sin2 2x sin2 2y ≥ 6

(sin2 x cos2 x+ sin2 y cos2 y

).

Mihály Bencze

PP. 19055. If x ∈ R, then(sin4 x+ cos2 x

) (cos4 x+ sin2 x

)≥ 13 .

Mihály Bencze

PP. 19056. If x ∈ R, then 12+cos2 x+cos4 x

+ 16+sin2 x

+ 11+sin4 x

≤ 1.

Mihály Bencze

PP. 19057. Solve in (0,+∞) the following system:x1√

1+8x1+ 13√x2 +

1√8+x3

=

= x2√1+8x2

+ 13√x3 +1√

8+x4= ... = xn√

1+8xn+ 13√x1 +

1√8+x2

= 1.

Mihály Bencze

PP. 19058. If x ∈ R, then 2+sin2 x1+cos6 x

+ 2+cos2 x

1+sin6 x≥ 269 .

Mihály Bencze


PP. 19058. If x ∈ R, then 11+sin2 x cos2 x

+ 11+2 sin2 x

+ 11+2 cos2 x

≥ 3.

Mihály Bencze

PP. 19059. If x ∈(0, π2

), then 5 (sinx+ cosx) + 3

√2

sin 2x ≥ 18− 5√2.

Mihály Bencze

PP. 19060. If x ∈ R, then 2+sin2 x3+cos4 x

+ 2+cos2 x

3+sin4 x≥ 1914 .

Mihály Bencze

PP. 19061. If x ∈[0, π2

], then

1+4√2+(2+

√2)(sinx+cosx)

(2+sinx)(2+cosx) +sin+x cosx

2+√2

≤ 2.

Mihály Bencze

PP. 19062. If x ∈[0, π2

], then 1sinx+cosx +

2+sin2 x√2+sinx

+ 2+cos2 x√

2+cosx≥ 3.

Mihály Bencze

PP. 19063. If x ∈ R, then 12 +(9√2− 7

)sinx cosx ≥ 7

√2 (sinx+ cosx) .

Mihály Bencze

PP. 19064. If x, y ∈(0, π2

)and t > 0 then

(t+sinx+sin y)5

sinx sin y +(t+cosx+cos y)5

cosx cos y ≥ 2t(1 + t2

).

Mihály Bencze

PP. 19065. If x ∈ R, then 3√

(sinx cosx)8 + 3√16(

3√sin8 x+

3√cos8 x

)≤ 3.

Mihály Bencze

PP. 19066. If x ∈ R, then2+2 sin2 x cos2 x

(1+2 sin2 x cos4 x)(1+2 cos2 x sin4 x)+ 5(1+4 sin2 x)(1+4 cos2 x)

+ 9(1+8 sin2 x)(1+8 cos2 x)

≥ 2.

Mihály Bencze

PP. 19067. If x ∈ R, then 424−sin2 2x +

52(2+sin2 x)(2+cos2 x)

≤ 35 .

Mihály Bencze


PP. 19068. If x ∈(0, π2

)then 5

(sinx+ cosx+

√2)+ 3

√2

sin 2x ≥ 18.

Mihály Bencze

PP. 19069. If x, y ∈(0, π2

)then

(sinx+ cosx) (1− sinx cosx) + (sin y + cos y) (1− sin y cos y) ≤ 2√2.

Mihály Bencze

PP. 19070. Solve the following system:(x21 + 1

)(x1 + 1)

2 + 7x22 = 5x3(x24 + x4 + 1

)(x22 + 1

)(x2 + 1)

2 + 7x23 = 5x4(x25 + x5 + 1

)−−−−−−−−−−−−−−−−−−−−(x2n + 1

)(xn + 1)

2 + 7x21 = 5x2(x23 + x3 + 1

) .Mihály Bencze


x31+1

(x1+1)3 +

5x22(x2+1)

2 =x32+1

(x2+1)3 +

5x32(x3+1)

2 = ... =x3n+1

(xn+1)3 +

5x12(x1+1)

2 =78 .

Mihály Bencze

PP. 19072. Solve the following system: 3√2x1 + 1 +

3√4− x2 + 3

√3 =

= 3√2x2 + 1 +

3√4− x3 + 3

√3 = ... = 3

√2xn + 1 +

3√4− x1 + 3

√3 = 0.

Mihály Bencze

PP. 19073. If ak > 0 (k = 1, 2, ..., n) andn∑k=1

a5k =15√3

, then∑cyclic

a1a2a35√a101 + a

102 + a

103 ≤ 1.

Mihály Bencze

PP. 19074. If xk > 0 (k = 1, 2, ..., n), then∏cyclic

1+x2(1+x21)(1+x1x2)1+x1x2+x21x2

≥

(1 + n

√n∏k=1

xk + n

√n∏k=1

x2k

)nMihály Bencze


PP. 19075. Solve the following system:2(2x41 + 2x

22 − 1

)= 3

(x23 + x4 − 1

)2(2x42 + 2x

23 − 1

)= 3

(x24 + x5 − 1

)−−−−−−−−−−−−−−−−2(2x4n + 2x

21 − 1

)= 3

(x22 + x3 − 1

) .Mihály Bencze

PP. 19076. Solve the following system:2|x1+2| =

∣∣2x2+1 − 1∣∣+ 2x3+1 + 12|x2+2| =

∣∣2x3+1 − 1∣∣+ 2x4+1 + 1−−−−−−−−−−−−−−2|xn+2| =

∣∣2x1+1 − 1∣∣+ 2x2+1 + 1.

Mihály Bencze

PP. 19077. Solve the following system:√3 sin 2x1 = 2 cos

2 x2 + 2√2 + 2 cos 2x3√

3 sin 2x2 = 2 cos2 x3 + 2

√2 + 2 cos 2x4

−−−−−−−−−−−−−−−−−√3 sin 2xn = 2 cos

2 x1 + 2√2 + 2 cos 2x2

.

Mihály Bencze

PP. 19078. Solve the following system:32x1+1 = 3x2+2 +

√1− 6 · 3x3 + 32(x4+1)

32x2+1 = 3x3+2 +√

1− 6 · 3x4 + 32(x5+1)−−−−−−−−−−−−−−−−−−32xn+1 = 3x1+2 +

√1− 6 · 3x2 + 32(x3+1)

.

Mihály Bencze

PP. 19079. If xn =

√√√√2 +√2 + ...+√2︸︷︷︸n−time

, thenn∏k=1

(4xk+1 − xk) < 6n.

Mihály Bencze


PP. 19080. Prove that for all n ∈ N∗ exist a natural number x such thatthe sum of digits of x2 is 19n.

Mihály Bencze

PP. 19081. In all triangle ABC holds(2a2 + 2bc− b2 − c2

) (2b2 + 2ca− c2 − a2

) (2c2 + 2ab− a2 − b2

)≥ 256Rs2r3.

Mihály Bencze

PP. 19082. Let be a, b, c > 0 such that λabc > a3 + b3 + c3. If λ ∈ (0, 5]then a, b, c are the sides of a triangle.

Mihály Bencze

PP. 19083. Prove that for all n ∈ N∗ exist a natural number x such thatthe sum of digits of x2 is 4n.

Mihály Bencze

PP. 19084. Solve the following system:x21 − 20x2 + 20 = 3x3−1x22 − 20x3 + 20 = 3x4−1−−−−−−−−−−x2n − 20x1 + 20 = 3x2−1

.

Mihály Bencze


1). 5s2 ≥ 27r2 + 54Rr2). 4s2 ≥ 27Rr3). 27s2Rr

(s2 + r2 + 4Rr

)≤ 54s2R2r2 +

(s2 + r2 + 4Rr

)34). 27Rs2 ≤ 2 (4R+ r)35). 27 (2R− r)

(s2 + r2 − 8Rr

)≤ 54Rr2 + 32 (2R− r)3

6). 27 (4R+ r)(s2 + (4R+ r)2

)≤ 54Rs2 + 32 (4R+ r)3

Mihály Bencze


PP. 19086. If ak > 0 (k = 1, 2, ..., n) and λ ≥ 1 then∑cyclic

aλ+11aλ2+a

λ3+...+a

λn≥ nn−1

√√√√√ n∑k=1 aλ+1kn∑k=1

aλ−1k

.

Mihály Bencze

PP. 19087. If ak > 0 (k = 1, 2, ..., n) then

nn−2(

n∏k=1

ak

)(n∑

k=1

1ak

)(

n∑k=1

ak

)n−1 + n∑k=1

ak ≥ (n+ 1) n+1√

n∏k=1

ak.

Mihály Bencze

PP. 19088. If ak > 0 (k = 1, 2, ..., n) and λ ≥ 0 then∑cyclic

aλ+11aλ2+...+a

λn≥

nn∑

k=1aλ+1k

(n−1)n∑

k=1aλk

Mihály Bencze

PP. 19089. If x, y ∈[0, π2

]then

(sinx

√1 + cos2 y + sin y

√1 + cos2 x

)2+

+(cosx

√1 + sin2 y + cos y

√1 + sin2 x

)2≤ 6− 2 cos (x− y) .

Mihály Bencze

PP. 19090. If xk ∈ [−1, 1] (k = 1, 2, ..., n), then

∑cyclic

√1− x1x2 +

√(1− x21

) (1− x22

)≤

√√√√2(n2 − ( n∑k=1

xk

)2).

Mihály Bencze

PP. 19091. If x ∈ R, then∣∣∣∣ n∑k=1

{kx} − n {x}∣∣∣∣ ≤ n− ln (n+ 1) when {·}

denote the fractional part.

Mihály Bencze


PP. 19092. Let ABC be a triangle and O denote his circumcentre. IfAO ∩BC = {D} , BO ∩AC = {E} , CO ∩AB = {F} , then9R2 ≤ AD +BE + CF ≤

9R2

4r .

Mihály Bencze

PP. 19093. If x, y ∈ (0, 1) then x(y2+1)

y(y+1)2+

y(x2+1)x(x+1)2

≥ 1−x1+x +1−y1+y .

Mihály Bencze

PP. 19094. Solve the following system:(5x21 + 5x1 − 8

) (21x22 + 49x2 + 22

)=(13x23 + 27x3 + 7

)2(5x22 + 5x2 − 8

) (21x23 + 49x3 + 22

)=(13x24 + 27x4 + 7

)2−−−−−−−−−−−−−−−−−−−−−−−−−(5x2n + 5xn − 8

) (21x21 + 49x1 + 22

)=(13x22 + 27x2 + 7

)2 .Mihály Bencze


1 + ctg2A2 ≥ 6.

Mihály Bencze

PP. 19096. If ak, bk > 0 (k = 1, 2, ..., n) andn∏k=1

ak ≥n∑k=1

a1a2...ak−1bkak+1...an, thenn∑k=1

ak ≥(

n∑k=1

√bk

)2(A

generalization of problem L.IX.214 Revista de matematica din Valea Jiului).

Mihály Bencze

PP. 19097. Solve the following system:(sinx1 +

√1 + sin2 x1

) (cosx2 +

√1 + cos2 x2

)=

=(sinx2 +

√1 + sin2 x2

) (cosx3 +

√1 + cos2 x3

)= ...

=(sinxn +

√1 + sin2 xn

) (cosx1 +

√1 + cos2 x1

)= 1.

Mihály Bencze

PP. 19098. Solve in Z the equation x2 + y2 + z2 + t2 = u2 + v2 + w2.

Mihály Bencze


PP. 19099. Let A1A2...An be a convex polygon and M ∈ Int (A1A2...An) .Denote B1, B2, ..., Bn the midpoints of A1M,A2M, ..., AnM. Prove thatA1B2, A2B3, ..., AnB1 are the sides of a convex polygon if and only if M isthe controid of the polygon A1A2...An.

Mihály Bencze

PP. 19100. Determine all ak ∈ Z (k = 1, 2, ..., n) , (a1, a2, ..., an) = 1 forwhich sin akx = 0 (k = 1, 2, ..., n) .

Mihály Bencze

PP. 19101. Prove thatn∏k=1

(710 −

(1

k+1 +1

k+2 + ...+12k

))≤ 4−nn! .

Mihály Bencze

PP. 19102. Let ABCD be a tetrahedron and M ∈ Int (ABCD) . DenoteA1, B1, C1, D1 the midpoint of AM,BM,CM,DM. Determne all points Mfor which AB1, BC1, CD1, DA1 are the sides of a convex quadrilateral.

Mihály Bencze

PP. 19103. In all triangle ABC holds∏ (1−ctgA2 ctgB2 )2

1+ctgA2ctgB

2

≥ 1.

Mihály Bencze

PP. 19104. Determine all functions f : N∗ → N∗ for which

f (x+ y) = f (x) + f (y) + 3 (x+ y) 3√f (x) f (y) for all x, y ∈ N∗.

Mihály Bencze

PP. 19105. Solve the following system:(x21 − 3x1 + 4

) (5x22 − 7x2 + 11

)= 5

(x23 − x3 + 1

)2(x22 − 3x2 + 4

) (5x23 − 7x3 + 11

)= 5

(x24 − x4 + 1

)2−−−−−−−−−−−−−−−−−−−−−−(x2n − 3xn + 4

) (5x21 − 7x1 + 11

)= 5

(x22 − x2 + 1

)2 .Mihály Bencze


PP. 19106. 1). Prove that the equation(x2 − xy + y2

) (z2 + zt+ t2

)= u2

have infinitely many solutions in Z

2). Solve the given equation in Z

Mihály Bencze

PP. 19107. Let ak ∈ {0, 1, ..., 9} (k = 1, 2, ..., n) for which a1a2...a2n,a1a2...an, an+1an+2...a2n are perfect p power.

Determine all r, n, p ∈ N for which a1a2...a2n can be expressed like a sum ofr perfect p powers.

Mihály Bencze

PP. 19108. Let ABC be a triangle and D ∈ (BC), E ∈ (CA) , F ∈ (AB) .

Prove that 1 +∑√

1 + AD2

BD·DC ≥s2+r2

2Rr .

Mihály Bencze

PP. 19109. If x > 0 and a > b > 0 then xa − xb ≥(ba

) aa−b −

(ba

) ba−b .

Mihály Bencze

PP. 19110. If a, b, c ∈ (0, 1) or a, b, c > 1 and x, y, z > 0 and y + z ≥ 2xthen logaxbycz a+ logbxcyaz b+ logcxaybz c ≥ 3x+y+z (A generalization ofproblem L. 859, Cardinal, Vol. XXI, Nr. 2, 2010/2011).

Mihály Bencze

PP. 19111. Prove that x1 = sinπ48 and x2 = cos

π48 are roots of the equation

65536x16+163840x12−33768x10+103424x8−40960x6+6576x4−256x2+1 = 0.

Mihály Bencze



512x8 − 1024x6 + 640x4 − 128x2 + 3−√5 = 0.

Mihály Bencze



x41 + 1 = 2x2

(2x23 + 7x3 + 2

)x42 + 1 = 2x3

(2x24 + 7x4 + 2

)−−−−−−−−−−−−x4n + 1 = 2x1

(2x22 + 7x2 + 2

) .Mihály Bencze



256x8 − 512x6 + 304x4 − 48x2 + 1 = 0.

Mihály Bencze


π24 are solutions of the

equation 256x8 − 512x6 + 320x4 − 64x2 + 1 = 0.

Mihály Bencze

PP. 19116. Solve the following system:2(√

x1 + 1 +√x22 + 6x2 + 1

)= 2x23 + 7x3 + 4

2(√

x2 + 1 +√x23 + 6x3 + 1

)= 2x24 + 7x4 + 4

−−−−−−−−−−−−−−−−−−−−2(√

xn + 1 +√x21 + 6x1 + 1

)= 2x22 + 7x2 + 4

.

Mihály Bencze

PP. 19117. If a > b > 0 and x ≥ 1, then

a(√x+ 2− 3

√x+ 2

)≥ (a− b)

(√x+ 1− 3

√x+ 1

)+ b (

√x− 3

√x) .

Mihály Bencze


loga1((n− 2) an−11 + a2a2...an

)≥ n (n− 1) + n2

logn−1

(n∏

k=1ak

) (Ageneralization of problem L. 882, Cardinal, XXI, Nr. 2, 2010/2011).

Mihály Bencze


PP. 19119. Solve in (0,+∞) the following system:

√x1 +

29 =

3√x2√

x2 +29 =

3√x3

−−−−−−√xn +

29 =

3√x1

.

Mihály Bencze

PP. 19120. If α > β > 0 then solve in (0,+∞) the following system:

xα1 − xβ2 = x

α2 − x

β3 = ... = x

αn − x

β1 =

(βα

) αα−β −

(βα

) βα−β

.

Mihály Bencze

PP. 19121. Solve in Z the equationn∏k=1

(x2k + xk + 1

)= 6y.

Mihály Bencze

PP. 19122. Solve in C the following system: zn−11 + (n− 1) |z2z3...zn| =

zn−12 + (n− 1) |z3z4...z1| = ... = zn−1n + (n− 1) |z1z2...zn−1| =n∑k=1

zn−1k .

Mihály Bencze

PP. 19123. Solve in (0,+∞) the following system:n∑k=1

xk = n

n∏k=1

loga+1 (ax+ 1) = 1when a ≥ 1.

Mihály Bencze

PP. 19124. If a, b, c > 0 thena+b+c

3 ≥3√abc+ 34

∣∣∣( 3√a− 3√b)( 3√b− 3√c) ( 3√c− 3√a)∣∣∣ .Mihály Bencze

PP. 19125. 1). If a, b > 0, A (a, b) = a+b2 , G (a, b) =√ab then for all

x, y ∈ (a, b) holds a < A(a,b)G2(x,y)−2G2(a,b)A(x,y)+G2(a,b)A(a,b)

G2(x,y)−2A(a,b)A(x,y)+2A2(a,b)−G2(a,b) < b.

2). Generalize for all xk ∈ (a, b) (k = 1, 2, ..., n) .

Mihály Bencze


PP. 19126. 1). If a, b, c ∈ R and abc 6= 0 then the equationsax2 + cx+ b = 0; bx2 + ax+ c = 0, cx2 + bx+ a = 0, have a common rootsthen 2 |a+ b+ c| ≤ |a|+ |b|+ |c| .

2). What happens when a, b, c ∈ C∗?3). Generalization.

Mihály Bencze

PP. 19127. If x ∈[0, π2

]then

(1 + 2 sin2 x

) (1 + cos2 x

)·

·(sinx+ cosx+ 1√

2

)2≥ 94 (sinx+ cosx)

2(

1√2+ sinx

)2 (1√2+ cosx

)2.

When holds the equality?

Mihály Bencze

PP. 19128. Solve in R the equation: (arctgx)3 +(arctg 1x

)3= 63π

3

864 .

Mihály Bencze

PP. 19129. Prove that∞∑k=1

((k2+2k2+1

)3− 1)< π

2

2 .

Mihály Bencze

PP. 19130. If x ∈[0, π2

]then

(cosx+√2 sin2 x) sinx

1+√2 cosx

+(sinx+

√2 cos2 x) cosx

1+√2 sinx

+ 1+sin 2x2√2(sinx+cosx)

≥

≥ (1+2√2 cos3 x) sinx

2(cosx+√2 sin2 x)

+(1+2

√2 sin3 x) cosx

2(sinx+√2 cos2 x)

+√2(sinx+cosx)(1−sinx cosx)

1+sin 2x .

When holds the equality?

Mihály Bencze

PP. 19131. If x, y, z > 0, then 3√∑

x3

3xyz +3

√∑x3y3

3x2y2z2+√∑

xy∑x2

+√

xyz∑x∑

x2y2≥ 4.

Mihály Bencze

PP. 19132. If x, y, z > 0, then∑ yz

x+y3+z2≤∑x+∑xy+3xyz

(∑x)2

.

Mihály Bencze


PP. 19133. Solve in (0,+∞) the following system:

2n∑k=1

2xk = 2n+2

2n∏k=1

log2 xk = 1

.

Mihály Bencze

PP. 19134. Compute∫ x(x+sinx+cosx)dx

(x2−1) sinx cosx+x(cos 2x+cosx+sinx)+cosx−sinx+1 for all

x ∈(0, π2

).

Mihály Bencze

PP. 19135. Determine all f : R→ R for whichf (f (x+ 2y) + f (2x+ y)) = f (f (x+ y) + 2 (x+ y)) for all x, y ∈ R.

Mihály Bencze

PP. 19136. If x, y, z > 0, then(∑

(x+ y) z3) (∑ x+y

z2

)≥ 4xyz (

∑x)(∑ 1

x

).

Mihály Bencze

PP. 19137. If a, b, c > 0 and∑ab = 1, then 40

√3∑ 1

1+a2≤ 81 (

∑a+ abc) .

Mihály Bencze

PP. 19138. Solve in N the following equation:(x2n + (xy)n + y2n

) (y2n + (yz)n + z2n

) (z2n + (zx)n + x2n

)= (xyz)n+1 .

Mihály Bencze

PP. 19139. Solve in C the following system:(x1 + 2) (x2 + 3) (x3 − 4) (x4 − 5) = (x2 + 2) (x3 + 3) (x4 − 4) (x − 5) = ... =(xn + 2) (x1 + 3) (x2 − 4) (x3 − 5) = 44.

Mihály Bencze

PP. 19140. Solve in R the following system:(1 + 3x1) (1 + 5x2) ≤ 9x23 + 10x4 + 5(1 + 3x2) (1 + 5x3) ≤ 9x24 + 10x5 + 5−−−−−−−−−−−−−−−−(1 + 3xn) (1 + 5x1) ≤ 9x22 + 10x3 + 5

.

Mihály Bencze


PP. 19141. In all triangle ABC holds 5√3

9 ≤2sr

s2−(2R+r)2 < 2.

Mihály Bencze

PP. 19142. Solve in R the following system:x21 − x2 + [x3] = x22 − x3 + [x4] = ... = x2n − x1 + [x2] = 2, when [·] denote theinteger part.

Mihály Bencze

PP. 19143. If a, b, c > 0 then∑ a

a2+ab+bc+ca≤ 94(a+b+c) .

Mihály Bencze

PP. 19144. Solve the following system:x+ y + z = 6x3 + y3 + z3 = 3

(x2 + y2 + z2

)+ 6

x5 + y5 + z5 + 20(x2 + y2 + z2

)= 5(x4 + y4 + z4) + 66

.

Mihály Bencze

PP. 19145. If ak ∈ C (k = 1, 2, ..., n) then determine all z ∈ C for which|z − a1 + a2 + ...+ an−1| = |(n− 1) z + a1 − a2 + ...+ an−1||z − a2 + a3 + ...+ an| = |(n− 1) z + a2 − a3 + ...+ an|− − −−−−−−−−−−−−−−−−−−−−−−−−|z − an + a1 + ...+ an−2| = |(n− 1) z + an − a1 + ...+ an−2|

.

Mihály Bencze

PP. 19146. 1). If z ∈ C and λ ∈ (0,+∞] ∪ [1,+∞) , thenn∑k=1

|z − k|λ ≥ n1−2λ(n2−14

)λ2). If λ ∈ (0, 1), then

n∑k=1

|z − k|λ ≤ n(n−12

)λ.

Mihály Bencze

PP. 19147. If x ∈(0, π2

)then

1). 2sinx+cosx+

√2 sin 2x

≥(3− sinx√

cosx− cosx4√2

)·(3− cosx√

sinx− sinx4√2

)2). sinx4√2+

√cosx

+ cosx4√2+√sinx

+√2√

sinx+√cosx

≥ 32 .

Mihály Bencze


PP. 19148. If x ∈(0, π2

)then

1). sinx3+2 cos2 x

+ cosx3+2 sin2 x

≤ 3−sin 2x4√2

2). 19+2 sin2 x−2

√2 cosx

+ 19+2 cos2 x−2

√2 sinx

+ 12(5−sin 2x) ≤38

Mihály Bencze

PP. 19149. If x ∈(0, π2

)then(

tgx+ cosx√2

+√2

sinx

)(ctgx+ sinx√

2+

√2

cosx

)≥ 81

3+2√2(sinx+cosx)+sin 2x

Mihály Bencze

PP. 19150. If x ∈(0, π2

)then

1). 42 +(18

√2− 13

)sin 2x ≥ 26

√2 (sinx+ cosx)

2). 15−sin 2x +10−2

√2(sinx+cosx)

25−2√2(sinx+cosx)+4 sin 2x

≤ 13). (2− sinx cosx)

(√2− sinx

) (√2− cosx

)≥ 12

Mihály Bencze

PP. 19151. Solve in R the following system:(x1 − 1)2 e2x1−2 + 4x2ex2−1 + x23 + 1 = 2

(x4 +

(x25 + 1

)ex5−1

)(x2 − 1)2 e2x2−2 + 4x3ex3−1 + x24 + 1 = 2

(x5 +

(x26 + 1

)ex6−1

)−−−−−−−−−−−−−−−−−−−−−−−−−−−(xn − 1)2 e2xn−2 + 4x1ex1−1 + x22 + 1 = 2

(x3 +

(x24 + 1

)ex4−1

) .Mihály Bencze

PP. 19152. If a, b, c > 0 and abc = 1 then∑ 2+(b2+3b+4a)c

c(b+1)(b+3) ≥154 .

Mihály Bencze

PP. 19153. If x ∈ R, then(sin4 x+ cos2 x

) (cos4 x+ sin2 x

) (1 + sin2 x cos2 x

)≤ 1.

Mihály Bencze

PP. 19154. If a, b, c > 0 and abc = 1 then

1).∑√a(b+c)

ab+1 ≥ 2

2).∑√ a+c

c(a+1) ≥ 2

Mihály Bencze


PP. 19155. If a, b, c > 0 and abc = 1 then

1).∑√ a

(3a+1)c ≥32

2).∑√ ab

bc+3 ≥32

Mihály Bencze

PP. 19156. Let A1A2...An be a convex polygon. In exterior of sides, and insame direction we consider the points B1, B2, ..., Bn such thatA2B1 = A2A3;A3B2 = A3A4, ..., AnB2 = A1A2. Compute

Area[B1B2...Bn]Area[A1A2...An]

.

Mihály Bencze

PP. 19157. If x ∈ R then1). 28

(1 + 3 sin2 x

) (1 + 3 cos2 x

)≤ 125 + 262 sin2 x cos2 x

2). 52(1 + 3 sin4 x

) (1 + 3 cos4 x

)≤ 255 + 4 sin4 x cos4 x

Mihály Bencze

PP. 19158. Solve in R the following system:1

1+x21− 21+x2x3 +

11+x24

= 11+x22

− 21+x3x4 +1

1+x25= ... = 1

1+x2n− 21+x1x2 +

11+x23

= 0.

Mihály Bencze

PP. 19159. Determine all a, b, c > 0 for which∑ a

2a2+bc≥ 3∑ a .

Mihály Bencze

PP. 19160. 1). Prove that exist infinite arithmetic progression, whichcontains an infinite geometric progression.2). Prove that exist infinite geometric progression, which contains an infinitearithmetic progression.

Mihály Bencze

PP. 19161. If ak ∈ R (k = 1, 2, ..., n) and λ ≥ 2, then solve the followingequations:

1).n∑k=1

sin(x+ak)λk−1

= 0

2).n∑k=1

cos(x+ak)λk−1

= 0

Mihály Bencze



2sR + 2

R+rR +

∑2sinA+sinB+cosC +

∑2c

octogon mathematical magazine, vol. 19, no.2, october ...432 octogon mathematical magazine, vol. 19,...

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