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204 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
Proposed problems
PP24615. 24 Let a, b, c, d,α be a numbers from (0, 1) interval,f : (0, 1) → R a convex and decreasing function. Ten are true the followinginequality f
�1− a3
�+ f�1− b3
�+ f�1− c3
�+ f�1− d3
�≥
≥ f�1− a2b
�+ f�1− b2c
�+ f�1− c2a
�+ f�1− d2a
�.
Marius Dragan
PP24616. Find the best k ∈ Z such that�√n+
√n+ 1 +
√n+ 2 +
√n+ 3
�=�√
16n+ k�for each n ∈ N.
Marius Dragan
PP24617. Let x, y, z be an interior numbers. Find all the rest dividing thenumber x3yz + xy3z + xyz3 by 11.
Marius Dragan
PP24618. Let a, b, c be the positive numbers such that a = b+ c. then istrue the following inequality:�1 + 1
a2
�a2 �1− 1
a2−b2−c2
�≤�1 + 1
b2
�b2 �1 + 1
c2
�c2.
Marius Dragan
PP24619. Let a, b, c, d be a positive real numbers such that a ≤ b ≤ c. Thenis true the following inequality:�a2 + 14
� �b2 + 14
� �c2 + 14
�≥ 26 (3a+ 2b+ c+ 1)2 .
Marius Dragan
PP24620. Let A,B be two square matrices from C, a, b, c, d a strict naturalnumbers such that a < b, q the quotiend of dividing of b to a, such thatcq �= d and AaBc = AbBd = In. Then it exist a strictly integer number suchthat Bu = In.
Marius Dragan
24Solution should be mailed to editor until 30.12.2018. No problem is ever permanently
closed. The editor is always pleased to consider for publication new solutions or new in sights
on past problems.
Proposed Problems 205
PP24621. Let n be a positive integer. Prov that the equation
22n �
x8 + y8�2n
= z2 + t2 + w2 has the integer solutions.
Marius Dragan
PP24622. We consider the complex number Z such that |z| = 1. Re z ≥ 0,Im z > 0. If we denote x = π
2 arg z and we have��z[x] − 1��+��z[3x] − z[2x]
�� =��z[2x] − z[x]
��+��z[3x] − 1
�� . Then√2 ≥��z[x] − 1
�� ·��z[3x] − z[2x]
�� ·��z[2x] − z[x]
�� ·��z[3x] − 1
�� ≥ 1.
Marius Dragan
PP24623. Let be f : R → (0,+∞) be an increasing concave function. Prove
thatn�
k=1
f�
1(2k+1)
√2k−1
�≤ nf
�1n
�.
Mihaly Bencze
PP24624. Compute� � � dxdydz
(242x5−(y−1)5−(z+1)5)(242y5−(z−1)5−(x+1)5)(242z5−(x−1)5−(y+1)5)where
x, y, z > 12 .
Mihaly Bencze and Gyorgy Szollosy
PP24625. Solve in C the following system�x2 − 9
�(y − 2) (z + y) +
�y2 − 36
�(z − 4) (x+ 8) =
=�y9 − 9
�(z − 2) (x+ 4) +
�z2 − 36
�(x− 4) (y + 8) =
=�z9 − 9
�(x− 2) (y + 4) +
�x2 − 36
�(y − 4) (z + 8) = −153.
Mihaly Bencze and Ferenc Olosz
PP24626. If ak > 0 (k = 1, 2, ..., n) then
n+��a1
a2
�n≥� a1a2...an−1
an−1n
+� an−1
1a2a3...an
.
Mihaly Bencze
PP24627. If x0 ∈ R and xn+1
�2x2n − xn + 3
�= 3x2n − xn + 2 for all n ∈ N,
then compute limn→∞
n (1− xn) .
Mihaly Bencze
PP24628. Compute F (m) =∞�n=0
�n!m!(n+m)!(2n)!(2m)!
�2.
Mihaly Bencze
206 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24629. If a ≥ 1 and f, g : [0, a] → [0, 1] are two increasing functions and
g is continuous, then compute limn→∞
n
�ag (0)−
a�ag (fn (x)) dx
�.
Mihaly Bencze
PP24630. If a, b, c > 0, thenab4 + b3c2 + a5c+ 3a2b2c ≥ c
�a2b2 + b3c+ a3c
�+
3√a2b2c2
�a3 + abc+ b3
�.
Mihaly Bencze
PP24631. If ak > 0 (k = 1, 2, ..., n) , then
3n�
k=1
a2k +�
cyclic
a21(a1+a2)4(a1+a3)
4+a22(a2+a1)4(a2+a3)
4+a23(a3+a1)4(a3+a2)
2
(a1+a2)4(a2+a3)
4(a3+a1)4 ≥ 3n
2 .
Mihaly Bencze
PP24632. If x > 1 then π2
6 > 1336 +
∞�n=2
�1− x−1
xn+1−1− (n−1)x
n+1
�2.
Mihaly Bencze
PP24633. If zk ∈ C∗ (k = 1, 2, ..., n) , thenn�
k=1
���zk + 1zk
���8≥ 16n
�1 + 2
n
n�k=1
Re�z2k��2
.
Mihaly Bencze
PP24634. Determine all x, y, z ∈ C for which3An = 2n (A+ I3) + (−1)n (2I3 −A) for all n ∈ N∗, where
A =
0 x yy 0 zz x 0
.
Mihaly Bencze
PP24635. Prove that 14
�π2
6 − 1�<
∞�n=2
�1�0
xndxx2+1
�2
< π2
24 .
Mihaly Bencze
Proposed Problems 207
PP24636. If a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , then one of equationsx3 − abcx2 + bcdx− cda = 0,x3 − bcdx2 + cdax− dab = 0,x3 − cdax2 + dabx− abc = 0,x3 − dabx2 + abcx− bcd = 0,have real solutions.
Mihaly Bencze
PP24637. In all scalene triangle ABC holds
1).� r2a
(ra−rb)2(ra−rc)
2 ≥ 1(4R+r)2−2s2
2).� AI4
(AI2−BI2)2(AI2−CI2)≥ 1
s2+r2−8Rr
Mihaly Bencze
PP24638. Let ABC be a triangle. Determine all n, p, k ∈ N for which��
a�
ama
�n+�
�
a�
ama
�p+�
�
a�
ama
�k≥
≥�
aamn
a+bmpb+cmk
c+
�
a
ampa+bmk
b+cmnc+
�
aamk
a+bmnb +cmp
c.
Mihaly Bencze
PP24639. If zk, wk ∈ C (k = 1, 2, ..., n) , n ≥ 2 and|z1| = |z2| = ... = |zn| = |w1| = |w2| = ... = |wn| ∈ (0, 1] then determine all
x, y ∈ R for which 2
����n�
k=1
zx+y2
k −n�
k=1
wx+y2
k
���� ≤n�
k=1
��zxk − wyk
��+n�
k=1
��zyk − wxk
�� .
Mihaly Bencze
PP24640. Let be (xn)n≥1 an increasing positive real numbers sequence and
λ = limn→∞
xn. Compute limn→∞
n
�λ− λ−xn
ln λxn
�.
Mihaly Bencze
PP24641. Let ABC be a triangle. Determine all λ ≥ 1 for which�
(sinA)λ
�
(cos Aλ )
λ ≥�
sinA�
cos Aλ
.
Mihaly Bencze
208 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24642. Solve in R the equationn�
k=1
�x+kk+1
�= 2m, when [·] denote the
integer part.
Mihaly Bencze
PP24643. If ak > 0 (k = 1, 2, ..., n) , then
n
n�
k=1
1ak
n�
k=1
1ak
≥n�
k=1
a1akk .
Mihaly Bencze
PP24644. Determine all p ∈ N for which (pn)! =n�
k=1
� pk
pk−1
�pn−k
.
Mihaly Bencze
PP24645. Solve in R the following system:
4 arctan 1−x11+x2
= π − 4 (arctanx3)2
4 arctan 1−x21+x3
= π − 4 (arctanx4)2
−−−−−−−−−−−−−−−−4 arctan 1−xn
1+x1= π − 4 (arctanx2)
2
.
Mihaly Bencze
PP24646. Solve in C the following system:x61 − 8x52 + 18x43 − 6x34 − 12x25 + 2x6 = x62 − 8x53 + 18x44 − 6x35 − 12x26 + 2x7 =... = x6n − 8x51 + 18x42 − 6x33 − 12x24 + 2x5 = −1.
Mihaly Bencze
PP24647. Denote A (n) =Lpn−Ln
p when p is a prime and Ln denote the nth
Lucas numbers. Compute∞�n=1
11+A2(n)
.
Mihaly Bencze
PP24648. Solve in N the equationn�
k=1
(kr + 1) = pr, where r ∈ Z.
Mihaly Bencze
Proposed Problems 209
PP24649. In all triangle ABC holds 35 ≤� ln(ma)
ln(mam2bmc)
< 1.
Mihaly Bencze
PP24650. If S (n, p) =n�
k=1
kp then
t�r=1
�2t+12r
�S (n, 2r) = 1
2
�(n+ 1)2t+1 + n2t+1 − 2n− 1
�.
Mihaly Bencze
PP24651. Compute limn→∞
n
�lnΓ (x+ 1) + lnΓ (1− x)−
n�k=1
ζ (2k) x2k
k
�,
when x ∈ (−1, 1) .
Mihaly Bencze
PP24652. If x ∈�0, 12�, then
cosx (sinx)2n−1 − sinx (cosx)2n−1 ≥ 2x��
2x2�n−1 −
�1− 2x2
�n−1�for all
n ∈ N∗.
Mihaly Bencze
PP24653. In all triangle ABC holds256sRr ≤ (a+ 2s) (b+ 2s) (c+ 2s) ≤ 4sR(R+2r)
r2.
Mihaly Bencze
PP24654. Compute limn→∞
n
�ζ (λ)−
n�i=1
n�j=1
(i−1)!(j−1)!(i+j)!
�.
Mihaly Bencze
PP24655. In all triangle ABC holds�
cos2�
tgA2tgB
2
1−tgA2tgB
2
�> 3
2 .
Mihaly Bencze
PP24656. In all triangle ABC holds 272 ≤� 1
tgA2tgB
2 (1−tgA2tgB
2 )≤ s2
2r2.
Mihaly Bencze
210 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24657. Compute limn→∞
n
�3e4
256 −n�
k=2
4e2
�1 + 1
k
�2k+1 (k−1)(k+1)(2k−1)(2k+1)
�.
Mihaly Bencze
PP24658. If a, b, c ∈ C∗ then 6561�|a|2 + |b|2 + |c|2
�−1·
·�
1|a2| +
1|b2| +
1|c2|
�−1· |abc|2 ≤
�3 |a+ b+ c|2 +� |2a− b− c|2
�·
·�3 |ab+ bc+ ca|2 +� |2bc− ac− ab|2
�.
Mihaly Bencze
PP24659. Determine all λ ∈ R for which
����n�
k=0
(−1)kλ
���� ≤�nλ�when [·]
denote the integer part.
Mihaly Bencze
PP24660. If a, b, c > 0 then� √
a+√b
a+b+c+3√ab
≥ 1√a+b+c
.
Mihaly Bencze
PP24661. In all triangle ABC holds� (tgA
2 )x
(tgA2 )
y+(tgB
2 )y ≥ 3
2
�1√3
�x−yfor all
x > y ≥ 0.
Mihaly Bencze
PP24662. Compute1�0
lnk�x
1k − (1− x)
1k
�dx when k ∈ N∗
Ovidiu Furdui and Mihaly Bencze
PP24663. Compute
limn→∞
n
�arctan
�tan�
π√2
�√2− 1
�cth�
π√2
�√2 + 1
�− π
8 −n�
k=1
arctan�
1k2+1
���.
Mihaly Bencze
PP24664. Compute limn→∞
n
�1−
n�k=1
kk−1e−k
k!
�.
Mihaly Bencze
Proposed Problems 211
PP24665. If x > y > 0 and a > 1 then�a−1a
�2loga
xy < x−y
xy .
Mihaly Bencze
PP24666. If P2 = 4�2−
√2, P3 = 8
�2−�2 +
√2,
P4 = 16
�2−�2 +�2 +
√2 etc, then compute
limn→∞
n�π3
8 − 4n (Pn − Pn−1)�.
Mihaly Bencze
PP24667. In all triangle ABC holds�� a+b+c
−a+b+c
� 2λa−a+b+c ≥ s2
3λ−1r2for all
λ ≥ 1.
Mihaly Bencze
PP24668. If a > 1 thenn(an+1−1)
n+1 > n(n+1)4
�a−1a
�n+
n�k=1
ak−1k .
Mihaly Bencze
PP24669. Compute∞�0
dxx+ex+e2x
.
Mihaly Bencze
PP24670. Compute
√3�
0
ln(1+√3x)dx
cos2 x.
Mihaly Bencze
PP24671. Compute limn→∞
n�1 + (n+1)n+1
nn − (n+2)n+2
(n+1)n+1
�.
Mihaly Bencze
PP24672. Compute Sk =1�0
dxxk(1+lnx)
.
Mihaly Bencze
212 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24673. Compute S (k, p) =n�
i=1
�nk
pi
�where [·] denote the integer part.
Mihaly Bencze
PP24674. Compute
√22�0
(1−x2) lnxdx√x(1−2x2)
.
Mihaly Bencze
PP24675. If x0 = 0 and xn+1 = λxn + xn−1 for all n ≥ 1, then determine allλ ∈ R for which xn =
�n1
�+ λ�n3
�+ λ2
�n5
�+ λ3
�n7
�+ ...
Mihaly Bencze
PP24676. Compute1�0
ln�
1+x1+
√1−x2
�√1−x2
dx.
Mihaly Bencze
PP24677. If P0 (x) = 1 and Pn (x) =dn
2nn!·dxn
�x2 − 1
�nthen
n�k=1
cos2 (kx)P 2n−k (cosx) ≥ n2P 2
n (cosx) for all n,m ∈ N, m < n and all
x ∈ R.
Mihaly Bencze
PP24678. Compute∞�n=1
nn
(n!en)2.
Mihaly Bencze
PP24679. Computeb�a
ln(b+1−x)dxln(b+1−x)+ln(x−a+1) where 0 < a < b.
Mihaly Bencze
PP24680. Determine all λ ∈ R for which limn→∞
n�k=0
1n+λk > 1
λ .
Mihaly Bencze
Proposed Problems 213
PP24681. Let a, b, c ∈ N∗ and x1 = a, xn+1 = xn + bn+ c. Compute∞�n=0
1xn+1
.
Mihaly Bencze
PP24682. If λ = limn→∞
nlnn
n−1�k=1
1k(n−k) , then compute
limn→∞
n
�λ− n
lnn
n−1�k=1
1k(n−k)
�.
Mihaly Bencze
PP24683. In all acute triangle ABC holds�s2+r2+Rr
2sr
�2+�
s2+r2−4R2
s2−(2R+r)2
�2≥ 12 + 4R
r + 8R(R+r)
s2−(2R+r)2.
Mihaly Bencze
PP24684. Let p ≥ 3 be a prime. Determine all r ∈ N for whichp−1�k=0
�kp−1k
�n
is divisible by p if n is even and not divisible by p if n is odd.
Mihaly Bencze
PP24685. Determine all p ∈ N for whichn�
k=1
(−1)k−1 kp = (−1)n Tm when
Tm is a triangular number.
Mihaly Bencze
PP24686. In all triangle ABC holds� (a+b)2
c2≤ 4sR
r .
Mihaly Bencze
PP24687. In all triangle ABC holds� cos 2A+cos 2B
sinC ≤ s2+r2
sr + 2�2− 3
√3�.
Mihaly Bencze
PP24688. Determine all prime p for which L22n ≡ (p− 1)mod p and
L22n+1 = − (p− 1)mod p when Ln is the nth Lucas number.
Mihaly Bencze
214 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24689. If z, w �= 1 and zn = wn = 1, then��(n− 2) z2 − 2 (n− 1)w + n��+��(n− 2)w2 − 2 (n− 1) z + n
�� ≤≤ (n−1)(2n−1)
6
�|z − 1|3 + |w − 1|3
�for all n ∈ N, n ≥ 2.
Mihaly Bencze and Jose Luis Dıaz-Barrero
PP24690. Solve in R the following system:
sinx1 + cosx2 = sinx2 + cosx3 = ... = sinxn = cosx1 =12
�3 +
√5.
Mihaly Bencze
PP24691. If sinx+ sin y = k (cos y − cosx)2 for x, y ∈ R, then determine allk ∈ R for which sin
�k2x�+ sin
�k2y�= 0.
Mihaly Bencze
PP24692. Determine all a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for
�abcd�bcda
+�bcda�cdab
+�cdab�dabc
+�dabc�abcd
is a perfect square.
Mihaly Bencze
PP24693. Compute Ak,p =∞�n=1
(−1)n−1 nk
pn .
Mihaly Bencze
PP24694. Determine all xk ∈ R (k = 1, ..., n) for which
[x1]3x2+{x3} + {x2}
3x3+[x4]≥ 4
15[x2]
3x3+{x4} + {x3}3x4+[x5]
≥ 415
−−−−−−−−−−−−[xn]
3x1+{x2} + {x1}3x2+[x3]
≥ 415
, when [·] denote the integer part, and {·}
denote the fractional part.
Mihaly Bencze and Jose Luis Dıaz-Barrero
PP24695. Compute� (cosx−sin2 x) cosxdx
sinx cosx(sinx−cosx)+(1+cos x) cosx .
Mihaly Bencze
Proposed Problems 215
PP24696. If Φ = 1+√5
2 then determine all a, b ∈ R for which the series∞�n=1
1na|sin(nbπΦ)| converges.
Mihaly Bencze
PP24697. Find all polynomials P for whichP (x− 2)P (x− 1)P (x) = P
�x3�for all x ∈ R.
Mihaly Bencze
PP24698. Determine all a, b, c, d ∈ N for which F2an+2b − 2cn− 2d isdivisible by 5 for all n ∈ N , when Fk denote the kth Fibonacci number.
Mihaly Bencze
PP24699. Denote xa, xb, xc the distance of orthocenter H to the sides
BC,CA and AB respectively. Prove that xa + xb + xc ≤ 6
�9s2R3r√
3.
Mihaly Bencze
PP24700. Compute∞�n=1
(p (n− 1) + k)xn−1 if |x| < 1 and p, k ∈ Z.
Mihaly Bencze
PP24701. Find the volume of the region bounded by the surfacex2n+1 + y2n+1 + z2n+1 = (xyz)n when n ∈ N∗.
Mihaly Bencze
PP24702. Determine all k,m ∈ N for which the sum of (m+ k) (kn+ 1)consecutive terms of the Fibonacci sequence is divisible by the(kn+m+ 1)st Lucas number.
Mihaly Bencze
PP24703. In all triangle ABC holds:2(s2−r2−Rr)s2+r2+2Rr
+ 3 3√4Rsr4s ≥ 2.
Mihaly Bencze
216 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24704. Evaluate the sumn�
k=0
(−1)k�nk
�3�3n3k
�−1.
Mihaly Bencze
PP24705. If x ∈�0, π2�then 1 < 1
sin2 x cos2 x+ 1
ln(sin2 x)+ 1
ln(cos2)x< 2.
Mihaly Bencze
PP24706. If x ∈ (0, 1) then n2 < xn−1
(x−1)xn +n�
k=1
1ln(1−xk)
< n.
Mihaly Bencze
PP24707. If x0 = 0, xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1. Determine all
λ ∈ R for which 1 ≤n�
k=1
xk
(1+x0+x1+...+xk)λ(xk+xk+1+...+xn)
λ ≤ π2 .
Mihaly Bencze
PP24708. If ak ∈ (0, 1] (k = 1, 2, ..., n) , then�
cyclic
1a21+a22
≥ 2[n2 ]+1
[n2 ]+1+n�
k=1ak
.
Mihaly Bencze
PP24709. Let be 1kλ
+ 1(k+1)λ
+ ...+ 1nλ = p
q when k, n,λ ∈ N∗. Determine
all λ ∈ N∗ for which p is odd.
Mihaly Bencze
PP24710. If x, y, z > 0 then
(x+ y + z)�
1x + 1
y + 1z
�≥ 3 + 2 (x+ y + z)
�(xxyyzz)
1x+y+z
xyz .
Mihaly Bencze
PP24711. Computeπ�0
sin2n+1 x sin (kx) dx, when n, k ∈ N.
Mihaly Bencze
Proposed Problems 217
PP24712. Let Z × Z × Z → R be a function which satisfies f (0, 0, 0) = 1and f (n, p, k) + f (n+ 1, p, k) + f (n, p+ 1, k) + f (n, p, k + 1) +f (n+ 1, p+ 1, k) + f (n+ 1, p, k + 1)++f (n, p+ 1, k + 1) + f (n+ 1, p+ 1, k + 1) = 0 for all n, p, k ∈ Z. Prove that|f (n, p, k)| ≥ 1
4 for infinitely many integers n, p, k.
Mihaly Bencze
PP24713. Let Tn =�n+12
�for all n ≥ 1, be the nth triangular number. Show
that for all integers k ≥ 2, the sequence�T kn
�n≥1
does not contain anyinfinite subsequence with all terms in geometric progression.
Mihaly Bencze
PP24714. Solve in N the equation�nk
��n+1k+1
��n+2k+2
�=�3n+33k+3
�.
Mihaly Bencze
PP24715. Find the closed form of�p
�1 + 2
p2+ 4
p4
�−1when the product is
over all primes.
Mihaly Bencze
PP24716. Let ABCD be an equilateral tetrahedron with edge length dinscribed in a sphere. Let P be a point of minor semisphere ABC. LetPA = a, PB = b, PC = c. Is it possible for a, b, c and d to all be distinctpositive integers?
Mihaly Bencze
PP24717. The unsigned Stirling number of the first kind�nk
�is the number
of permutation of [n] that have k cycles. Express (−1)n�k
�n3k
�(−1)k ;
(−1)n�k
�n
3k+1
�(−1)k , (−1)n
�k
�n
3k+2
�(−1)k in function of combinatorial
expressions.
Mihaly Bencze
218 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24718. If λ = limn→∞
n1�0
�12
�nxdx, then compute
limn→∞
n
�λ− n
1�0
�12
�nxdx
�.
Mihaly Bencze
PP24719. Solve in Z the following equation(x+ 3)
�x2 + 3
� �x3 + 3
�= y2 + 15.
Mihaly Bencze
PP24720. Determine all a, b ∈ R for which1�0
(ax+ b)2k+1 dx =1�0
(ax+ b)2p+1 dx when k, p ∈ N.
Mihaly Bencze
PP24721. Compute
limn→∞
n
�2√3π9 − 1−
n�k=1
1(b−a)2n+1
b�a(x− a)n (b− x)n dx
�when 0 < a < b and
n ∈ N.
Mihaly Bencze
PP24722. If z1, z2, z3 ∈ C such that |z1| = |z2| = |z3| = a andz1 + z2 + z3 = b. Determine all a ∈ R and b ∈ C such that
|z − z1|2 + |z − z2|2 + |z − z3|2 = 3�1 + |z|2
�for all z ∈ C.
Mihaly Bencze
PP24723. If ak > 0 (k = 1, 2, ..., n) , thenn�
k=1
ak
ak+nn
n−1≤ n
n+1 .
Mihaly Bencze
PP24724. Solve in R the following systemarctg3x1 · arctg3x2 = arctg3x2 · arctg3x3 = ... = arctg3xn · arctg3x1 = π2
18 .
Mihaly Bencze
Proposed Problems 219
PP24725. Solve in R the following system
�√2 + 2 sinx = 2
54 sin y +
�√2− 2 sin z�√
2 + 2 sin y = 254 sin z +
�√2− 2 sinx�√
2 + 2 sin z = 254 sinx+
�√2− 2 sin y
Ionel Tudor and Mihaly Bencze
PP24726. Solve in N the equation(a+ b)b+c + (b+ c)a+b = (a+ b+ c)a+b+c .
Mihaly Bencze
PP24727. If xk > 1 (k = 1, 2, ..., n) andn�
k=1
xk = n, then
�logx1
�xn−12 + xn−2
2 + ...+ x2 + 1�≥ n(3n−1)
2
Mihaly Bencze
PP24728. Solve in N the equation ab + bc + ca = ba! + cb! + ac!.
Mihaly Bencze
PP24729. If xk > 1√n(k = 1, 2, ..., n) , then
�lognx1x2
�(n− 1)x1 +
x2x1
�≥ 1 + n(n−1)
log3
�
3n�
n�
k=1xk
�n−1� .
Mihaly Bencze
PP24730. If A,B ∈ Mn (Z) and AB = BA, detA = detB = 0, thendetermine all n ∈ N for which det (An +Bn) = xn + yn when x, y ∈ Z.
Mihaly Bencze
PP24731. If x, y, z > 0 and xyz = 1 then determine all λ ∈ R for which� 1+xλn
1+xn ≥ 1 for all n ∈ N.
Mihaly Bencze
PP24732. If ak > 0 (k = 1, 2, ..., n) , then
1n
m�i=1
ai1+ai2+...+ainn√
ai1ai2...a
in
+nm n
√a1a2...an
a1+a2+...+an≥ 2m.
Mihaly Bencze
220 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24733. Let A1A2...An be a convex polygon and S =n�
k=1
ak. Prove that
�n
�S−a2−a3S−2a1
≥ n.
Mihaly Bencze
PP24734. If a, b, c > 0 and ab +
bc +
ca ≤ 11 then
ab+ bc+ ca+ (a+ b+ c)�√
ab+√bc+
√ca�≤ 6 (a+ b+ c)2 .
Mihaly Bencze
PP24735. Solve in (0,+∞) the following
system
(1 +√x)
y= 1 + 2y
√z + x2�
1 +√y�z
= 1 + 2z√x+ y2
(1 +√z)
x= 1 + 2x
√y + z2
.
Mihaly Bencze
PP24736. Solve the following system
x4 + 12y2 + 6 = 6�z2 + 2t
�
y4 + 12z2 + 6 = 6�t2 + 2x
�
z4 + 12t2 + 6 = 6�x2 + 2y
�
t4 + 12x2 + 6 = 6�y2 + 2z
�
Mihaly Bencze and Ionel Tudor
PP24737. In all triangle ABC holds
1).� ma
ha≤
√3(s2−r2−4Rr)
2sr
2).��ma
ha
�2≤ (s2−r2−4Rr)
2
4s2r2
Mihaly Bencze
PP24738. Solve in R the following system:729x+6 ·27y+8 ·27−z = 729y+6 ·27z+8 ·27−x = 729z+6 ·27x+8 ·27−y = 15.
Mihaly Bencze
PP24739. In all triangle ABC holds1).� tgA
2+tg2A≤ R+r
R√3
2).� ctgA
2+ctg2A≤ s
R√3
Mihaly Bencze
Proposed Problems 221
PP24740. In all triangle ABC holds
1).� tg2A
(2+tg2A)(2tg2A+1)≤ s
3R
2).� ctg2A
(2+ctg2A)(2ctg2A+1)≤ R+r
3R
Mihaly Bencze
PP24741. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
1chxk
= 3 then� sh2xk
2+sh4xk≤ 1.
Mihaly Bencze
PP24742. Prove thatn�
k=1
k2(k+1)2
(k2+2)2(k2+2k+3)2≤ n
18(n+1) .
Mihaly Bencze
PP24743. Compute∞�
n1,n2,...,nk=1
1n21+n2
2+...+n2k.
Mihaly Bencze
PP24744. Solve in R the following system:
xy2+2
≤ 1√3(z+1)
yz2+2
≤ 1√3(x+1)
zx2+2
≤ 1√3(y+1)
.
Mihaly Bencze
PP24745. If a, b, c > 0 and 2abc+�
ab ≤ 1 then� ab
(2a2+1)(2b2+1)≤ 1
3 .
Mihaly Bencze
PP24746. Prove that∞�n=0
�n
n2+2
�4< π2
36 .
Mihaly Bencze
PP24747. Solve in N the following system
a�b2 + 3c+ 2
�= x2 + y2
b�c2 + 3a+ 2
�= y2 + z2
c�a2 + 3b+ 2
�= z2 + x2
.
Mihaly Bencze
222 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24748. Let ABC be a triangle. Determine all M ∈ Int (ABC) for whichBMC∡−A∡; AMC∡−B∡; AMB∡− C∡ are the angle of a triangle.
Mihaly Bencze
PP24749. If f : R → R is convex, then�
af�
b1+a+ab
�≥ (a+ b+ c) f (1)
when a, b, c > 0 and abc = 1.
Mihaly Bencze
PP24750. Solve in Z the following system:|x− 2015|+ y2 − 3z = |y − 2015|+ z2 − 3x = |z − 2015|+ x2 − 3y = 4.
Mihaly Bencze
PP24751. Solve in N the equation x2 + y2 = 169n when n ∈ N.
Mihaly Bencze
PP24752. Determine all xk ∈ C (k = 1, 2, ..., n) such that� x1x2...xn−1
1+x1+x1x2+...+x1x2...xn−1= 1.
Mihaly Bencze
PP24753. Compute
� �3x2 − 2y + 5
� �3y2 − 2z + 5
� �3z2 − 2x+ 5
�lnxyzdxdydz.
Mihaly Bencze
PP24754. In all triangle ABC holds12sr ≤� (b+ c)wa ≤ 2s
�3 (s2 + r2 − 8Rr) ≤
√3�
a2.
Mihaly Bencze
PP24755. Determine all f : [0, 1] → (0,+∞) of class C1 for which�1�0
f (x) dx
�n
≥1�0
fn+1 (x) dx when n ∈ N.
Mihaly Bencze
Proposed Problems 223
PP24756. Let f : [0,+∞) → R be a continuous function. Compute
limn→∞
n
�1−
1�0
f (nx) dx
�if lim
x→∞f (x) = 1.
Mihaly Bencze
PP24757. Determine all functions f : [0, 1] → (0,+∞) of class C1 for which
f (e) = 1 and1�0
f2 (x) dx+1�0
dx(f ′(x))2
≤ 2.
Mihaly Bencze
PP24758. Determine all functions f : (0,+∞) → (0,+∞) for which
f ′ (x) f�1y
�= 1
z
f ′ (y) f�1z
�= 1
xf ′ (z) f
�1x
�= 1
y
, for all x, y, z ∈ (0,+∞) .
Mihaly Bencze
PP24759. If ai > 0 (i = 1, 2, ..., n) and k ∈ {1, 2, ..., n} thenn�
i=1ai
n
�
n�
i=1ai
+kn
�
i=1ai
�
cyclic(a1+a2+...+ak)
≥ n+ 1. (A generalization of problem 5.343
Mathematical Reflections).
Mihaly Bencze
PP24760. Prove that∞�n=0
5n�
35n+1−5·35n+4
�
7295n−2435
n−35n+1
= 12 .
Mihaly Bencze
PP24761. If a0 = 0, a1 = 1, a2 = 2, a3 = 6 andan+4 = 2an+3 + an+2 − 2an+1 − an for all n ≥ 0 then denote α (n) =
�ann2
�
when [·] denote the integer part. Compute∞�n=1
1α2(n)
.
Mihaly Bencze
PP24762. In all triangle ABC holds�(|a− c|+ |b− c|)2 ≤ 8
�s2 − 3r2 − 12Rr
�.
Mihaly Bencze
224 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24763. Compute limn→∞
n
�1− (n+ 1)!
�e−
n�k=0
1k!
��.
Mihaly Bencze
PP24764. Solve in R the following system
√2x − 1 +
√3y − 2x +
√2z+1 − 3y = 2x + 1√
2y − 1 +√3z − 2y +
√2x+1 − 3z = 2y + 1√
2z − 1 +√3x − 2z +
√2y+1 − 3x = 2z + 1√
2x − 1 +√3y − 2x +
√2z+1 − 3y = 2x + 1
.
Longaver Lajos and Mihaly Bencze
PP24765. Solve in R the following system(3x + 8 log3 y + 8z) (3y − 8 log3 z + 8x) == (3y + 8 log3 z + 8x) (3z − 8 log3 x+ 8y) == (3z + 8 log3 x+ 8y) (3x − 8 log3 y + 8z) = 121.
Mihaly Bencze and Bela Kovacs
PP24766. In all scalene triangle ABC holds1).� (b−a)rc
mb−ma< 2 (4R+ r)
2).� (b−a)hahb
mb−ma< 4s2r
R
Mihaly Bencze
PP24767. Compute� � (1−cos 2x−sin 2y)(1−cos 2y−sin 2x)dx
(1+sin 2x)(1+sin 2y) .
Mihaly Bencze
PP24768. In all triangle ABC holds� tgA
2tgB
2
9tg3 A2tg2 B
2+tgC
2
≤ s6r .
Mihaly Bencze
PP24769. Compute limn→∞
n
�ln 2−
π2�π6
(1−(sinx)2n)ctgxdx1+(sinx)2n
�.
Mihaly Bencze and Ionel Tudor
PP24770. Solve in R the following system:
�x2 − 2y
�+ 2 [z] = [t]2�
y2 − 2z�+ 2 [t] = [x]2�
z2 − 2t�+ 2 [x] = [y]2�
t2 − 2x�+ 2 [y] = [z]2�
x2 − 2y�+ 2 [z] = [t]2
,
Proposed Problems 225
when [·] denote the integer part.
Mihaly Bencze
PP24771. Determine the left-most digit of the decimal expansion of20172000.
Mihaly Bencze
PP24772. If ak ∈ [−3, 3] (k = 1, 2, ..., n) then determine the maximum of�cyclic
��a21 − a2a3 + 1�� .
Mihaly Bencze
PP24773. In all triangle ABC holds2r�5s2 + r2 + 4Rr
�≤ 3 (R+ r)
�s2 + r2 + 2Rr
�.
Mihaly Bencze
PP24774. In all triangle ABC holds� (ctgA
2 )6
ctg2 A2+ctg2 B
2
≥ s2
2r2.
Mihaly Bencze
PP24775. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1, then
n−1�
cyclicx1x2...xn−1
− 1n�
k=1xk
≤ n−2n .
Mihaly Bencze
PP24776. In all triangle ABC holds��3r
s
�3rs ctg
A2 ctg
A2 + 2
�3rs ctg
A2 + 2
�≤ 216.
Mihaly Bencze
PP24777. Prove that exist infinitely many triangles ABC such that
�� 1m
rba
��� 1mrc
b
��� 1mra
c
�≥ 1
34R+r−1
�R2r2
�4R+r.
Mihaly Bencze
226 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24778. Determine all a, b, c ≥ 3 for which(a−1)2(b−1)
(c−2)a−2 + (b−1)2(c−1)
(a−2)b−2 + (c−1)2(a−1)
(b−2)c−2 ≤ ab + bc + ca.
Mihaly Bencze
PP24779. Find all functions f : R → R such that f (0) ∈ Q andf�x+ f2
�y3�+ f3
�z2��
= f6 (x+ y + z) for all x, y, z ∈ R.
Mihaly Bencze
PP24780. Solve in N the equation (2a + b)�2b + a
�= a!b!.
Mihaly Bencze
PP24781. Find all functions f : R → R such that(f ◦ f) (x− y) f
�x2 + xy + y2
�= x3 − yf
�y2�for all x, y ∈ R.
Mihaly Bencze
PP24782. Compute limn→∞
n
2−
1�
0(x2−x−2)
ndx
1�
0
(4x2−2x−2)ndx
.
Mihaly Bencze
PP24783. Compute limn→∞
n
2−
1�
0(2x2−5x−1)
ndx
1�
0
(x2−4x−1)ndx
.
Mihaly Bencze
PP24784. Compute limn→∞
n
2−
n+1�
k=1k! csc
�
π
2k
�
n�
k=1k!scs
�
π
2k
� + 2n
.
Mihaly Bencze
PP24785. In all triangle ABC holds 1s2+r2+2Rr
≥ 18R2 .
Mihaly Bencze
Proposed Problems 227
PP24786. In all triangle ABC holds
� (tgA2 )
n−1(tgB
2 )n−2
tgC2
ctgA2+ctgB
2
≥ 22n−2·3n−1rs for all n ∈ N, n ≥ 2.
Mihaly Bencze
PP24787. In all triangle ABC holds� ctgA
2
1+9tg2 A2tg2 B
2
≥ 12 .
Mihaly Bencze
PP24788. Compute limn→∞
n
�lnπ − 3
2 −n�
k=2
�k2 ln
�1− 1
k2
�+ 1��
Ovidiu Furdul and Mihaly Bencze
PP24789. In all triangle ABC holds� a3
a2+ab+b2≥ 9Rr
s .
Mihaly Bencze
PP24790. Prove thatn�
k=1
�e2 + π2k
�< 2ne
n+π(π2−1)e(π−1) .
Mihaly Bencze
PP24791. Prove thatn�
k=1
�e2 +
�e+ 1
k
�2�< 2ne2n+
en(n+1)2 .
Mihaly Bencze
PP24792. In all triangle ABC holds� ma
a ≥ s2−4Rr−r2
2sR .
Mihaly Bencze
PP24793. In all triangle ABC holds��
tgA2 tg
B2
�4 ≥�rs
�2.
Mihaly Bencze
PP24794. In all triangle ABC holds�� s3+27r3(ctgA
2 )3
s2+9r2(ctgA2 )
2
�2
≥ 9r (4R+ r) .
Mihaly Bencze
228 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24795. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = n, then
n�k=1
�1+a3k1+a2k
�2≥
n�k=1
a2k.
Mihaly Bencze
PP24796. Determine all convex pentagon ABCDE for whichPerimeter (ACE) ≥ 2BD.
Mihaly Bencze
PP24797. Let a, b ∈ Q such that a1k + b
1k + (ab)
1k is also rational.
Determine all k ∈ N∗ for which a1k and b
1k must be rationals.
Mihaly Bencze
PP24798. Let be p ≥ k ≥ 3 prime numbers andA =
�k≤i1<i2<...<ik≤p−1
i1i2...ik. Determine all k ∈ N∗ for which A+ 1 is
divisible by p.
Mihaly Bencze
PP24799. Let ABC be an acute-angled triangle with AC �= BC, letM ∈ Int (ABC) and F the foot of the altitude through C. Furthermore, letP and K be the feet of the perpendiculars dropped from A and Brespectively to the extension of C. The line FM intersects the circumcircleof FPK a second time at L. Determine all M for which ML < MF.
Mihaly Bencze
PP24800. The first n primes are p1 = 2, p2 = 3, ... etc. Set K = pp11 pp22 ...ppnn .Find all positive integers a, b such that K
a(a+1)...(a+b−1) is even, andK
a(a−1)...(a+b−1) has exactly a (a+ 1) ... (a+ b− 1) divisors.
Mihaly Bencze
PP24801. Let (d) a line through point A, and ABC be a triangle. The line(d) intersects the line BC at P . Let Q,R be the symmetrical of P withrespect to the lines AB respectively AC. Determine all (d) for whichBC ⊥ QR.
Mihaly Bencze
Proposed Problems 229
PP24802. Find all non-constant polynomialsP (x) = xn + an−1x
n−1 + ...+ a1x+ a0 with rational coefficients whose rootsare exactly the numbers 1
a0, 1a1, ..., 1
an−1with the same multiplicity.
Mihaly Bencze
PP24803. Find all non-decreasing functions f : R → R for whichf�f�x3�+ y + f
�z2��
= x3 + f (y) + f2 (z) for all x, y, z ∈ R.
Mihaly Bencze
PP24804. Prove thatn−1�k=0
�n−1k
� �(x+ k)k−1 (y + n− k)n−k−1 + (y + k)k−1 (x+ n− k)n−k−1
�=
= (x+y)(x+y+n)n−1
xy for all x, y ∈ C∗.
Mihaly Bencze
PP24805. Solve in Z the equation (x+ y + z)�
1x+1 + 1
y+1 + 1z+1
�= 9.
Mihaly Bencze
PP24806. Consider a triangle ABC with an inscribed circle with centre Iand radius r. Let CA, CB and CC be circles internal to ABC, tangent to itssides and tangent to the inscribed circle with corresponding radii rA, rB andrC . Show that rλA + rλB + rλC ≥ rλ
3λ−1 for all λ ∈ (−∞, 0] ∪ [1,+∞) . Whathappent if λ ∈ (0, 1)?
Mihaly Bencze
PP24807. If xk > 0 (k = 1, 2, ..., n) then
�cyclic
x21
x2≥ �
cyclic
(x1 − x2)2 +
�n
n�k=1
x2k.
Mihaly Bencze
PP24808. Determine all a, b ∈ N for which for each number(n!)a + 2b; (n!)a + 3b; ...; (n!)a + nb there exist a prime divisor that does notdivide any other number from this set.
Mihaly Bencze
230 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24809. In all triangle ABC holds�
ar2a ≥ 6sr (2R− r) .
Mihaly Bencze
PP24810. Let H0 = 0 and Hn =n�
k=1
1k . Determine all p ∈ N∗ for which
n�k=1
(−1)n−k �nk
�pkHk = pHn −H�
np
�, when [·] denote the integer part.
Mihaly Bencze
PP24811. If ak ∈ R (k = 1, 2, ..., n) then2n�k=1
|sin ak|+2n�k=1
|cos ak|+����sin�
2n�k=1
ak
�����+����cos�
2n�k=1
ak
����� ≤ 2 (2n+ 1) .
Mihaly Bencze
PP24812. In all triangle ABC holds� |cos 3A| ≤ 3
��2 +
√2− s
R
�.
Mihaly Bencze
PP24813. Compute min�ab + bc + ca
�where a, b, c > 0 and abc = 1.
Mihaly Bencze
PP24814. Find all injective functions f : Z → Z that satisfy|f (x)− f (y)|+
��f2 (y)− f2 (z)��+��f3 (z)− f3 (x)
�� ≤≤ |x− y|+
��y2 − z2��+��z3 − x3
�� for all x, y, z ∈ Z.
Mihaly Bencze
PP24815. Denote d (n) the number of different divisors of n, and letan+1 = an + kd (n) where a1 ∈ N∗. Determine all k ∈ N∗ for which doesthere exist an a1 such that k consecutive numbers of the sequence are perfectk powers of natural numbers?
Mihaly Bencze
PP24816. Let be ai, bi, ci ∈ C (i = 1, 2, ..., n) , |ai| = |bi| = |ci| = 1
(i = 1, 2, ..., n) and a = 1n
n�i=1
ai, b =1n
n�i=1
bi, c =1n
n�i=1
ci and
Proposed Problems 231
di = abci + abic+ aibc− 2aibici. Prove thatn�
i=1|di| ≤ n
Mihaly Bencze
PP24817. Compute limn→∞
n
�1
128 −n�
k=2
k2+1k3(k2−1)4
�.
Mihaly Bencze
PP24818. Determine all prime numbers Pk (k = 1, 2, ..., n+ 1) for which
P 2n+1 =
n�k=1
P 2k .
Mihaly Bencze
PP24819. If a, b, c > 0 then 67 +� a+b
2(a+b)+3c ≥ 2� a+b+2c
5(a+b)+4c .
Mihaly Bencze
PP24820. If xk ∈ [−5, 3] (k = 1, 2, ..., n) , then�cyclic
√3x1 − 5x2 − x1x2 + 15 ≤ 4n. When holds the equality?
Mihaly Bencze and Bela Kovacs
PP24821. If λ ∈ [0, 1] thenn�
k=1
�1
(2k+1)√2k−1
�λ≤ n1−λ
�1− 1√
2n+1
�λ.
Mihaly Bencze
PP24822. If A,B ∈ Mn (R) such that AB = On. Determine all m ∈ N for
which det
�In +
m�k=1
�Ak +Bk
��≥ 0.
Mihaly Bencze
PP24823. Solve in R the following system:
3x1 + 2x2 = 9log2 x3 + x243x2 + 2x3 = 9log2 x4 + x25−−−−−−−−−−−3xn + 2x1 = 9log2 x2 + x23
Mihaly Bencze and Lajos Longaver
232 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24824. Solve in R the following system:
x212x22 + 2 log2 x3 = x4 + 2x5+2x6
x222x23 + 2 log2 x4 = x5 + 2x6+2x7
−−−−−−−−−−−−−−−x2n2
x21 + 2 log2 x2 = x3 + 2x4+2x5
Mihaly Bencze and Lajos Longaver
PP24825. Determine all f : R → R for which(x+ 1)2 f (x+ 1)− (x+ 2)2 f (−x− 1) = (x+ 1)
�2x2 + 6x+ 5
�for all
x ∈ R.
Mihaly Bencze
PP24826. Determine all n ∈ N∗ for which kn is a perfect k power,pn is aperfect p power and rn is a perfect r power, when k < p < r are positiveintegers.
Mihaly Bencze
PP24827. 1). Let ÷a1, a2, ..., an be an arithmetical progression. Determine
the inverse of the matrice A =
a1 a2 ... an−1 anan a1 ... an−2 an−1
− − − −− −−a2 a3 ... an a1
2). What happens when a1, a2, ..., an is a geometrical progression?
Mihaly Bencze
PP24828. If a = 1+√2
2 then determine all n, k ∈ N for which an + a−k ∈ N.
Mihaly Bencze
PP24829. Solve in R the following system
2 + tgx1 =√3 + 2 [tgx2]
2 + tgx2 =√3 + 2 [tgx3]
−−−−−−−−−−−−2 + tgxn =
√3 + 2 [tgx1]
when [·] denote the integer part.
Mihaly Bencze and Lajos Longaver
Proposed Problems 233
PP24830. Solve in R the following system
�x21 − x22 − 1
� �x22 − x23 + 1
�= 4x1x2�
x22 − x23 − 1� �
x23 − x24 + 1�= 4x2x3
−−−−−−−−−−−−−−−−�x2n − x21 − 1
� �x21 − x22 + 1
�= 4xnx1
.
Mihaly Bencze and Ferenc Olosz
PP24831. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = 1 then
√5 ≤
n�k=1
√2ak + 1 ≤
�n (n+ 2).
Mihaly Bencze
PP24832. Determine all p, n, k ∈ N for which the function
f (x) =
�xn sin
�x−k�− x−p cos
�x−k�if x ∈ (0, 1)
0 if x = 0have primitive
functions.
Mihaly Bencze
PP24833. If xk > 1 (k = 1, 2, ..., n) , then determine all n ∈ N for whichx21
x2−1 +x22
x3−1 + ...+ x2n
x1−1 ≥ 2n+1.
Mihaly Bencze
PP24834. If ak > 0 (k = 1, 2, ..., n) , thenn�
k=1
�a2k + ak + 1
�≥ 3n
n�k=1
ak.
Mihaly Bencze
PP24835. Prove that 2√6n
n+1 ≤n�
k=1
3( 23)
k+2( 3
2)k
k(k+1) ≤ 5nn+1 .
Mihaly Bencze
PP24836. If a, b, x > 0 then axm+n + b ≥ xn (m+ n)��
an
�n � bm
�m� 1n+m
for
all n, n ∈ N∗.
Mihaly Bencze
234 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24837. Prove thatn(n+1)
2 ≤n�
k=1
50k(k+1)2k−45ke(k2+k)k−11e2kk+1
25k(k+1)2k−25ke(k2+k)2−6e2kk+1≤ 11n(n+1)
12 .
Mihaly Bencze
PP24838. Prove that 2n9(n+3) ≤
n�k=1
k+1(k+2)(k+3)(k2+k+1)
≤ n3(n+3) .
Mihaly Bencze
PP24839. Prove that∞�k=1
k2+2k√k2+1
> π2
3 .
Mihaly Bencze
PP24840. Prove thatn�
k=1
k2(k4+2k3+3k2+2k2+2k+2)k2+k+1
≥ n(n+1)(2n+1)3 .
Mihaly Bencze
PP24841. Solve in R the following
system
√x2 + x− 1 +
�y − y2 + 1 = z2 − z + 2�
y2 + y − 1 +√z − z2 + 1 = x2 − x+ 2√
z2 + x− 1 +√x− x2 + 1 = y2 − y + 2
.
Mihaly Bencze
PP24842. Solve in R the following system
x2
3+√
9−y2+ 1
4(3−√9−z2)
= y2
3+√9−z2
+ 14(3−
√9−x2)
= z2
3+√9−x2
+ 1
4�
3−√
9−y2� = 1.
Mihaly Bencze
PP24843. Solve in R the following system
�x2 + 3
�4= 256 (2y − 1)�
y2 + 3�4
= 256 (2z − 1)�z2 + 3
�4= 256 (2x− 1)
.
Mihaly Bencze
Proposed Problems 235
PP24844. Solve in R the following system
√x2 + 1 +
�y3 − 1 = z2 + z + 1�
y2 + 1 +√z3 − 1 = x2 + x+ 1√
z2 + 1 +√x3 − 1 = y2 + y + 1
.
Mihaly Bencze
PP24845. Compute limn→∞
n
�1− 2
n�k=1
2k2−14k4+1
�.
Mihaly Bencze
PP24846. Solve in R the following system
x2 + 13y + 4 = 6 (z + 2)√t
y2 + 13z + 4 = 6 (t+ 2)√x
z2 + 13t+ 4 = 6 (x+ 2)√y
t2 + 13x+ 4 = 6 (y + 2)√z
.
Mihaly Bencze
PP24847. Solve in R the following system�x+ 2 + 2
√y + 1 +
�y + 2− 2
√z + 1 =
�y + 2 + 2
√z + 1+
+�z + 2− 2
√x+ 1 =
�z + 2 + 2
√x+ 1 +
�x+ 2− 2
√y + 1 = 2.
Mihaly Bencze
PP24848. Compute limn→∞
n
�23 −
n�k=2
k3−1k3+1
�.
Mihaly Bencze
PP24849. Solve in R the following system
√x+ 4 +
√y − 4 = 2
�z2 +
√t2 − 16− 6
�
√y + 4 +
√z − 4 = 2
�t2 +
√x2 − 16− 6
�
√z + 4 +
√t− 4 = 2
�x2 +
�y2 − 16− 6
�
√t+ 4 +
√x− 4 = 2
�y2 +
√z2 − 16− 6
�.
Mihaly Bencze
PP24850. Solve in R the following system1
x−√
y2−y− 1
y+√z2−x
= 1y−
√z2−z
− 1z+
√x2−x
= 1z−
√x2−x
− 1
x+√
y2−y=
√3.
Mihaly Bencze
236 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24851. Solve in R the following system
√x− 2 +
√4− y = z2 − 6z + 11√
y − 2 +√4− z = x2 − 6x+ 11√
z − 2 +√4− x = y2 − 6y + 11
.
Mihaly Bencze
PP24852. Solve in R the following system√x2 + 32−2 4
�y2 + 32 =
�y2 + 32−2 4
√z2 + 32 =
√z2 + 32−2 4
√x2 + 32 = 3.
Mihaly Bencze
PP24853. Solve in R the following system√x2 + 9−
�y2 − 7 =
�y2 + 9−
√z2 − 7 =
√z2 + 9−
√x2 − 7 = 2.
Mihaly Bencze
PP24854. Solve in R the following systemx2 + 3y + 4
√z2 + 3t− 6 = y2 + 3z + 4
√t2 + 3x− 6 =
= z2 + 3x+ 4�x2 + 3y − 6 = t2 + 3y + 4
�y2 + 3z − 6 = 28.
Mihaly Bencze
PP24855. Solve in R the following system�3x2 − 2y + 15 +
�3y2 − 2z + 8 =
�3y2 − 2z + 15 +
√3z2 − 2x+ 8 =
=√3z2 − 2x+ 15 +
�3x2 − 2y + 8 = 7.
Mihaly Bencze
PP24856. Solve in R the following system√15− x+
√3− y =
√15− y +
√3− z =
√15− z +
√3− x = 6.
Mihaly Bencze
PP24857. Solve in R the following system3x
y2−4y+1− 2y
z2+z+1= 3y
z2−4z+1− 2z
x2+x+1= 3z
x2−4x+1− 2x
y2+y+1= 8
3 .
Mihaly Bencze
PP24858. Solve in R the following system3x2−1
y + 5y3z2−z−1
= 3y2−1z + 5z
3x2−x−1= 3z2−1
x + 5x3y2−y−1
= 1198 .
Mihaly Bencze
Proposed Problems 237
PP24859. Solve in R the following system
(x+ 1)2 = (y − 3)3
(y + 1)2 = (z − 3)3
(z + 1)2 = (x− 3)3.
Mihaly Bencze
PP24860. Solve in R the following system2x
2y2−5y+3+ 13y
2z2+z+3= 2y
2z2−5z+3+ 13z
2x2+x−3= 2z
2x2−5x+3+ 13x
2y2+y−3= 6.
Mihaly Bencze
PP24861. Solve in R the following system
�3x2 + 5y + 8 = 1 +
�3y2 + 5z + 1�
3y2 + 5z + 8 = 1 +√3z2 + 5x+ 1√
3z2 + 5x+ 8 = 1 +�3x2 + 5y + 1
Mihaly Bencze
PP24862. Solve in R the following system
�x− 1 + 2
√y − 2 = 1 +
�y − 1− 2
√z − 2�
y − 1 + 2√z − 2 = 1 +
�z − 1− 2
√x− 2�
z − 1 + 2√x− 2 = 1 +
�x− 1− 2
√y − 2
.
Mihaly Bencze
PP24863. Solve in R the following system
xy2−y+1
+ 2yz2+z+1
= 1y
z2−z+1+ 2z
x2+x+1= 1
zx2−x+1
+ 2xy2+y+1
= 1
.
Mihaly Bencze
PP24864. Solve in C the following system
2x2 − 8y + 12 =�y2 − 4z − 6
�2
2y2 − 8z + 12 =�z2 − 4x− 6
�2
2z2 − 8x+ 12 =�x2 − 4y − 6
�2.
Mihaly Bencze
PP24865. Solve in Z∗\ {−1} the equation 1x(y2+1)
+ 1(x+1)(y+1) +
1y(x2+1)
= 1.
Mihaly Bencze
238 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24866. Solve in Z∗ the equation 1x + 1
y + 1z = 1
9 + 1xyz .
Mihaly Bencze
PP24867. If x ∈ R\ {−1, 1} thenn�
k=1
�x2+1−1x−1
�k+
n�k=1
�x2n+1+1
x+1
�k> 2− 1
2n−1 .
Mihaly Bencze
PP24868. Solve in Z∗ the equation 1x3 + 1
x2y+ 1
xy2+ 1
y3= 1.
Mihaly Bencze
PP24869. Solve in R the following system
x12 + y4 + 1 ≥ z9 + ty12 + z4 + 1 ≥ t9 + xz12 + t4 + 1 ≥ x9 + yt12 + x4 + 1 ≥ y9 + z
.
Mihaly Bencze
PP24870. Prove thatn�
k=1
2007
�k2014+k2008
k4+1≥ n(n+1)
2 .
Mihaly Bencze
PP24871. Prove thatn�
k=1
k12+k4+1k8+1
≥ n(n+1)2 .
Mihaly Bencze
PP24872. Solve in R the following system
2x4 + 1 ≥ 2y3 + z2
2y4 + 1 ≥ 2z3 + x2
2z4 + 1 ≥ 2x3 + y2.
Mihaly Bencze
PP24873. Prove thatn�
k=1
2k4+1k+2 ≥ n(n+1)(2n+1)
6 .
Mihaly Bencze
PP24874. Prove that∞�k=1
kk+1 ≥ π2
12 .
Mihaly Bencze
Proposed Problems 239
PP24875. Solve in R the following system
�2x3 ≤ y + 12y3 ≤ x+ 1
.
Mihaly Bencze
PP24876. Prove that n(n−1)2 ≤
n�k=1
4√k4 + k3 + k2 + k + 1 ≤ n(n+3)
2 .
Mihaly Bencze
PP24877. Prove that n(n+1)2 ≥
n�k=1
2m�2m (k − 1) + 1 when m ∈ N∗.
Mihaly Bencze
PP24878. Solve in R the following system
13x1 + 1
5x2 + 115x3 = 2
1+15x4 + 23x5+25x6 + 2
5x7+9x81
3x2 + 15x3 + 1
15x4 = 21+15x5 + 2
3x6+25x7 + 25x8+9x9
−−−−−−−−−−−−−−−−−−−−−1
3xn + 15x1 + 1
15x2 = 21+15x3 + 2
3x4+25x5 + 25x6+9x7
.
Mihaly Bencze
PP24879. Solve in C the following system
|z1 − 1|+ |z2 + 1| =√2 |z3 − 1|
|z2 − 1|+ |z3 + 1| =√2 |z4 − 1|
−−−−−−−−−−−−−−|zn − 1|+ |z1 + 1| =
√2 |z2 − 1|
.
Dorin Andrica and
Mihaly Bencze
PP24880. Solve in C the following system
x21 + x22 + 16x23 = 9x24 + 1x22 + x23 + 16x24 = 9x25 + 1−−−−−−−−−−−−x2n + x21 + 16x22 = 9x23 + 1
.
Mihaly Bencze
240 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24881. Solve in C the following system
2�x21 + 4 =
�x22 + 1 +
�x23 + 9
2�x22 + 4 =
�x23 + 1 +
�x24 + 9
−−−−−−−−−−−−−2�x2n + 4 =
�x21 + 1 +
�x22 + 9
.
Mihaly Bencze
PP24882. Solve in R the following system
x10 + 20y4 = 32 + 10z6 + 21t5
y10 + 20z4 = 32 + 10t6 + 21x5
z10 + 20t4 = 32 + 10x6 + 21y5
t10 + 20x4 = 32 + 10y6 + 21z5
.
Mihaly Bencze and Ionel Tudor
PP24883. Solve in R the following system
4 cos π2x =
√y + tg π
2z4 cos π
2y =√z + tg π
2x
4 cos π2z =
√x+ tg π
2y
Mihaly Bencze and Ionel Tudor
PP24884. Solve in R the following system
x6 + 3y5 + 3z = 1 + 5t3
y6 + 3z5 + 3t = 1 + 5x3
z6 + 3t5 + 3x = 1 + 5y3
t6 + 3x5 + 3y = 1 + 5z3
.
Mihaly Bencze and Ionel Tudor
PP24885. Let n ∈ N and a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ; a �= 0 such thatabcd+ cd = 2n − 20. Prove that n ≥ 10 and determine all n ∈ N and abcd forwhich holds the given equality.
Ionel Tudor
PP24886. Solve in R the equation64 cos6 x+ 96 cos5 x− 40 cos3 x+ 6 cosx− 1 = 0.
Ionel Tudor
PP24887. a). If n ∈ N, n ≥ 3 then exist an increasing arithmeticalprogression x1, x2, ..., xn ∈ (−1, 1) such that 2x1 + 2x2 + ...+ 2xn = n
Proposed Problems 241
b). If n = 3 then determine one arithmetical progression which verify a).
Ionel Tudor
PP24888. Prove that the function f : [8,+∞) → R wheref (x) = 4 cos π
2x − tg π2x −√
x is increasing. Solve in N the equation4 cos π
2n =√n+ tg π
2n .
Ionel Tudor
PP24889. If x ∈ R thenx8 + x7 + 3x6 + 5x5 + x4 − 10x3 + 12x2 − 8x+ 16 > 0. Solve in R theequation x10 − 10x6 − 21x5 + 20x4 − 32 = 0.
Ionel Tudor
PP24890. If�1 + 1
x
�x= e
�1−
∞�k=1
bk(x+1)k
�when x > 0 then compute
∞�k=1
11+b2k
.
Mihaly Bencze
PP24891. If�1 + 1
x
�x= e
�1−
∞�k=1
dk
( 1112
+x)k
�, then compute
∞�k=1
11+d2k
.
Mihaly Bencze
PP24892. If
An = 2n−1
8 0 0 04n 4n+ 8 4n 4n
n2 − n n2 − n n2 − 5n+ 8 n2 − 5n−n2 − 3n −n2 − 3n −n2 + n −n2 + n+ 8
, then
compute∞�n=1
11+(det(An))2
.
Mihaly Bencze
PP24893. Solve in R the following system
x+ [y] = {z}+ 54y + [z] = {x}+ 54z + [x] = {y}+ 54
, when [·]
and {·} denote the integer respective the fractional part.
Mihaly Bencze
242 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24894. Solve in R the following system
�2x+ 11− 6
�2 (y + 1) < z+9
2�2y + 11− 6
�2 (z + 1) < x+9
2�2z + 11− 6
�2 (x+ 1) < y+9
2
.
Mihaly Bencze
PP24895. If x ∈ R then
32 + sinx+ cosx < 2 sin2 x
1+sin2 x+ 2 cos2 x
1+cos2 x+�
1+sin4 x2 +
�1+cos4 x
2 .
Mihaly Bencze
PP24896. If x ∈ R then�
1+sin4 x2 +
�1+cos4 x
2 + sinx+ cosx < 3.
Mihaly Bencze
PP24897. If λ = limn→∞
�1 +
n�k=1
(4k−1)!(4k)!−(4k−2)!
�then compute
limn→∞
n
�λ− 1−
n�k=1
(4k−1)!(4k)!−(4k−2)!
�.
Mihaly Bencze
PP24898. If ak > 0 (k = 1, 2, ..., n) , thenn�
k=1
a2k ≥ �cyclic
a1a2 +1
2r+1
�(|a1 − a2|+ |a1 − a3|+ ...+ |a1 − ar|)2 when
r ∈ {2, 3, ..., n− 1} .
Mihaly Bencze
PP24899. Solve in C the following system
x4 + 12y2 + 32 = 12z2 + 16ty4 + 12z2 + 32 = 12t2 + 16xz4 + 12t2 + 32 = 12x2 + 16yt4 + 12x2 + 32 = 12y2 + 16z
.
Mihaly Bencze
Proposed Problems 243
PP24900. Solve in C the following system
x√3 + y +
√39− 3z = 4
√3 + t2
y√3 + z +
√39− 3t = 4
√3 + x2
z√3 + x+
√39− 3x = 4
�3 + y2
t√3 + y +
√39− 3y = 4
√3 + z2
Mihaly Bencze
PP24901. Solve in C the following equation
(3x − 2)2x + (4x − 2)
2x + ...+ (nx − 2)
2x = (n−2)(2n−3)(n−1)
6 .
Mihaly Bencze
PP24902. Solve in N the following system
(x!)3 = xy + y + z
(y!)3 = yz + z + x
(z!)3 = zx+ x+ y
.
Mihaly Bencze
PP24903. Determine all n ∈ N for which�
cyclic
n
�xn+xy
xn−xy+yn ≤ 3 n√2 for all
x, y, z > 0.
Mihaly Bencze
PP24904. Prove thatn�
k=1
�1√k+�
k2+12k2
�≤ n+ 1.
Mihaly Bencze
PP24905. Solve in R the following system
x3 + 5y = z2 − 5 + 4 3
�(5t2 + t− 5)2 + 6 3
√5x2 + x− 5
y3 + 5z = t2 − 5 + 4 3
�(5x2 + x− 5)2 + 6 3
�5y2 + y − 5
z3 + 5t = x2 − 5 + 4 3
�(5y2 + y − 5)2 + 6 3
√5z2z +−5
t3 + 5x = y2 − 5 + 4 3
�(5z2 + z − 5)2 + 6 3
√5t2 + t− 5
.
Mihaly Bencze
244 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24906. Prove that
3 <
�4 sin4 x1+sin4 x
+�2�1 + sin8 x
�+�
4 cos4 x1+cos4 x
+�2 (1 + cos8 x) for all x ∈ R.
Mihaly Bencze
PP24907. Let ABC be a triangle and M1,M2,M3 ∈ Int (ABC) . Provethat 2 (AM1 +AM2 +AM3) +BM1 +BM2 +BM3 +CM1 +CM2 +CM3 <
< 3 (2BC +AB +AC) .
Mihaly Bencze
PP24908. Let A1A2A3...An be a convex polygon for whichA1A2 = A2A3 = An−1An = AnA1. If A1∡ ≥ A2∡ ≥ ... ≥ An∡ then thepolygon is regular.
Mihaly Bencze
PP24909. In all acute triangle ABC holds
exp
�A�0
tgxx dx+
B�0
tgxx dx+
C�0
tgxx dx
�>�
π3
(π−2A)(π−2B)(π−2C)
� 2π.
Mihaly Bencze
PP24910. Solve in M2 (N) the following system:A2
1 − 8A2 + 7I2 = A22 − 8A3 + 7I2 = ... = A2
n − 8A1 + 7I2 = O2.
Mihaly Bencze
PP24911. Prove that 15 + 1
7 + ...+ 12n+3 < ln
�n+22 .
Mihaly Bencze
PP24912. Solve in R the following system
log2�x31 + 2x2
�= 1 + logx3
(2 + x4)log2�x32 + 2x3
�= 1 + logx4
(2 + x5)−−−−−−−−−−−−−−−−−log2�x3n + 2x1
�= 1 + logx2
(2 + x3)
Mihaly Bencze
Proposed Problems 245
PP24913. Solve in R the following system
log9 (4x1 + 1) = log4 (9
x2 − 1)log9 (4
x2 + 1) = log4 (9x3 − 1)
−−−−−−−−−−−−−log9 (4
xn + 1) = log4 (9x1 − 1)
.
Mihaly Bencze
PP24914. Solve in R the following system
3�x1
√x2 − 7
�x3�x4
√x5 = 6 12
�x6
√x7
3�x2
√x3 − 7
�x4�x5
√x6 = 6 12
�x7
√x8
−−−−−−−−−−−−−−−−−−3�xn
√x1 − 7
�x2�x3
√x4 = 6 12
�x5
√x6
.
Mihaly Bencze
PP24915. If a1 > 0 and�n+ a4n
�a2n+1 = na2n for all n ≥ 1 and λ = lim
n→∞an
then compute limn→∞
n (λ− an) .
Mihaly Bencze
PP24916. Solve in R the following system�3x1 +
4x2+1 + 8
�
2(1+x23)
��3x2 +
4x3+1 + 8
�
2(1+x24)
�=
=
�3x2 +
4x3+1 + 8
�
2(1+x24)
��3x3 +
4x4+1 + 8
�
2(1+x25)
�= ...
=
�3xn + 4
x1+1 + 8�
2(1+x22)
��3x1 +
4x2+1 + 8
�
2(1+x23)
�= 81.
Mihaly Bencze
PP24917. Compute limn→∞
n
�2e
π2 − n
�n�
k=1
�1 + n2
k2
��.
Mihaly Bencze
PP24918. Solve in R the following system[x1] +
�xk2�= [x2] +
�xk3�= ... = [xn] +
�xk1�= k when k ∈ N∗ is given, and [·]
246 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
denote the integer part.
Mihaly Bencze
PP24919. In all triangle ABC holds�1 + sin A
2
� �1 + sin B
2
� �1 + sin C
2
�≤ (a+3b−c)(b+3c−a)(c+3a−b)
8abc .
Mihaly Bencze
PP24920. If x0 = x1 = 1 and n (n+ 1)�xn+1
√xn−1 − xn
√xn�=
√xn−1xn
for all n ≥ 1 then compute limn→∞
n (4− xn) .
Mihaly Bencze
PP24921. Solve in R the following system:�x21+x2
22 + x3+x4
2 =
�x22+x2
32 + x4+x5
2 = ... =
�x2n+x2
12 + x2+x3
2 = 9.
Mihaly Bencze
PP24922. Let ABC be a triangle, N ∈ (BC) ,M ∈ (CA) ,K ∈ (AB) suchthat NAC∡ = α, MBA∡ = β, KCB∡ = γ. Prove that�3 (NC2 +MA2 +KB2) ≥� ab sinα
b sinα+c sin(A−α) .
Mihaly Bencze
PP24923. If λ = limn→∞
n�k=1
e1
n+k
n+k then compute limn→∞
n
�λ−
n�k=1
e1
n+k
n+k
�.
Mihaly Bencze
PP24924. Solve in R the following system:
log3 2 log2 x1 = log2 (x2 − 1)log3 2 log2 x2 = log2 (x3 − 1)−−−−−−−−−−−−−log3 2 log2 xn = log2 (x1 − 1)
.
Mihaly Bencze
PP24925. Prove that (n!)2 <n�
k=1
�1
arctg 1k
�2<�(n+1)(2n+1)
6 + 1�n
.
Mihaly Bencze
Proposed Problems 247
PP24926. If ak =
�1
(arctg 1k )
2
�when [·] denote the integer part, then
n�k=1
2k+1aka2k+1
= n(n+2)
(n+1)2.
Mihaly Bencze
PP24927. Let p ∈ {1, 2, ..., n− 1} a given number. Determine all n ∈ N for
which
�n�
k=1
�1
arctg 1√k
�2�= p+ n(n+1)
2 when [·] denote the integer part.
Mihaly Bencze
PP24928. If a, b,λ > 0 thenb�a
√lnxdx ≤ (λ−1)(b−a)
2√λ
+ 12√λln bb
aa .
Mihaly Bencze
PP24929. If x, y, z > 1 then�
log x+y2
�13
�x�≤ 27
�
(x+y)2
8xyz(�
x)3
�logx
x+y2 .
Mihaly Bencze
PP24930. If Fn denote the nth Fibonacci number, then�n�
k=1
1Fk
�2
≥n�
k=1
k+1F 2k.
Mihaly Bencze
PP24931. Prove thatn�
k=1
�arctg 1
kk
�2> n2
n2+1.
Mihaly Bencze
PP24932. If ak ∈ (0, 1) (k = 1, 2, ..., 2n+ 1) then
2n+1�k=1
logak
�1
2n+1
2n+1�k=1
ak
�≥
(2n+1)2n+12n+1�
k=1ak
�
2n+1�
k=1ak
�2n+1 .
Mihaly Bencze
248 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24933. If x > 0 (k = 1, 2, ..., n) andn�
k=1
xk = n, then
n�k=1
xn−1k +
n�k=1
xk + n− 1 ≥ 2�
x1x2...xn−1.
Mihaly Bencze
PP24934. If A,B ∈ Mn (C) such that xBA− yAB = In. Determine allx, y ∈ C for which det (AB −BA) = 0.
Mihaly Bencze
PP24935. If a > 1 and x, t > 0 then ax
x − ax+t
x+t ≤ tx(x+t) .
Mihaly Bencze
PP24936. In all triangle ABC holds�� rb+rc
ra
�λ≥ 3�Rr
�λfor all λ ≥ 1.
Mihaly Bencze
PP24937. Solve in R the following system:
(13− 7x1)�1 + 3
1x2
�= 16 (2 + 7x3)
(13− 7x2)�1 + 3
1x3
�= 16 (2 + 7x4)
−−−−−−−−−−−−−−−−−(13− 7xn)
�1 + 3
1x1
�= 16 (2 + 7x2)
Mihaly Bencze and Gyorgy Szollosy
PP24938. If x0, x1 > 1 and xn+1 = log3 (1 + xn+1 + xn) then computelimn→∞
n (1− xn) .
Mihaly Bencze
PP24939. In all triangle ABC holds� r3a
r2a+2rbrc≥ 4R+r
3 .
Mihaly Bencze
PP24940. In all triangle ABC holds� 1
(tgA2+tgB
2 )(tgB2+tgC
2 )≤ 9
4 .
Mihaly Bencze
Proposed Problems 249
PP24941. Solve in R the following system:
11a − 2b = c2
11b − 2c = a2
11c − 2a = b2
Gheorghe Stoica and Mihaly Bencze
PP24942. If b1 ≥ 2b2 ≥ ... ≥ nbn ≥ 0 then b1 +
�n�
k=1
√bk
�2
≥n�
k=1
(k + 1) bk.
Mihaly Bencze
PP24943. Let ABC be a triangle, A1 ∈ (BC) , B1 ∈ (CA) , C1 ∈ (AB) .The medians AA1, BB1, CC1 intersect second time the circumcircle in pointsD,E and F. Prove that
s2 + r2 + 4Rr +BD ·DC + CE ·AE +AF · FB ≤ AD2 +BE2 + CF 2.
Mihaly Bencze
PP24944. In all triangle ABC holds� r2a
r3b+r3c≥ 3(4R+r)
2(4R+r)2−4s2.
Mihaly Bencze
PP24945. Determine all n ∈ N for which
�n�
k=1
1
(arctg 1k )
2
�= n(n+1)(2n+1)
6 + p
when p ∈ {1, 2, ..., n− 1} is given.
Mihaly Bencze
PP24946. If p ∈ N∗ is given, then compute the integer part ofn�
k=1
p
�p+ 1
k! .
Mihaly Bencze
PP24947. Compute limn→∞
n
�e−1 −
n�p=1
pap
�n−1�
p=1(p+1)ap
where
ap = 1 +n�
k=1
f ′k (0) and fk (x) = kx (1− x) (2− x) ... (k − x) .
Mihaly Bencze
250 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24948. If x0 ∈ [0,π] and (n+ 1)xn =n�
k=0
sinxk then compute
limn→∞
n�√
3−√lnnxn+1
�.
Mihaly Bencze
PP24949. Solve in M3 (R) the equation x3 + x2 + x =
1 1 11 1 11 1 1
.
Mihaly Bencze
PP24950. If a ∈ (0, 1) ∪ (1,+∞) then solve in R the following system
2ax+1 = (a− 1)2 y2 +�a2 − 1
�z + 2a
2ay+1 = (a− 1)2 z2 +�a2 − 1
�x+ 2a
2az+1 = (a− 1)2 x2 +�a2 − 1
�y + 2a
Mihaly Bencze and Gyorgy Szollosy
PP24951. If x, y, z ∈ R and z + y + z = (2k + 1)π, k ∈ Z then determineall a, b, c ∈ R for which
�sin ax sin (by − cz) =
�sin ax (sin by − sin cz)
Mihaly Bencze
PP24952. In all triangle ABC holds�
(wa)b+c ≤
�√3(s2−r2−4Rr)
2s
�4s
.
Mihaly Bencze
PP24953. In all triangle ABC holds�� cos2 B
2+cos2 C
2
cos2 A2
≤�
2�(4R+r)3+s2(2R+r)
2Rs2+ 1�.
Mihaly Bencze
PP24954. In all triangle ABC holds� r2a
(4R+r)ra+3r2b≥ 1
2 .
Mihaly Bencze
PP24955. Determine all a, b ∈ N for which�7a + 51b − 2
� �7b + 51a − 2
�is
divisible by 64.
Mihaly Bencze
Proposed Problems 251
PP24956. Compute∞�n=0
(n!)a((n+2)!)b
((2n+3)!)cif a+ b = c; a, b, c ∈ N.
Mihaly Bencze
PP24957. Prove that�
n�k=0
�nk
�2ak�− 1
n
+
�n�
k=0
�nk
�1ak
�− 1n
=
[n2 ]�k=0
�n2k
��2kk
�ak
(1+n)2k
− 1n
for all a > 0
and n ∈ N∗.
Gyorgy Szollosy
PP24958. In all triangle ABC holds�� sin2 B
2+sin2 C
2
sin2 A2
≥√2 +�
(2R−r)(s2+r2−8Rr)−2Rr2
8Rr2.
Mihaly Bencze
PP24959. Solve in R the equation√60− x+
√65− x+
√156− x = 15.
Gyorgy Szollosy
PP24960. If a, b, c > 0 and a ≥ b+ c then solve the equation√bc+ x+
√ca+ x+
√ab+ x = 3a−b−c
2 .
Mihaly Bencze
PP24961. In all triangle ABC holds�� rb+rc
ra≤ 3�
Rr .
Mihaly Bencze
PP24962. If ak ∈ (0, 1) (k = 1, 2, ..., n− 1) such thatn−1�k=1
ak > 1 then solve
in R the following system:
ax11 + ax2
2 + ...+ axn−1
n−1 = (a1 + a2 + ...+ an)xn
ax21 + ax3
2 + ...+ axnn−1 = (a1 + a2 + ...+ an)
x1
−−−−−−−−−−−−−−−−−−−−axn1 + ax1
2 + ...+ axn−2
n−1 = (a1 + a2 + ...+ an)xn−1
.
Mihaly Bencze
252 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24963. If xk ≥ 1 (k = 1, 2, ..., n) , then
�cyclic
x71+6
x62−3x5
2+5x42−5x3
2+3x22−x2+1
≥ 7n.
Mihaly Bencze
PP24964. Solve in R the following system
arccos x12 + arccos 2x2 = arccos x2
2 + arccos 2x3 = ... =arccos xn
2 + arccos 2x1 =2π3 .
Mihaly Bencze
PP24965. If ak > 0 (k = 1, 2, ..., n) , then�
cyclic
a41+1
2a22−3a2+2≥ 2n n
�n�
k=1
ak.
Mihaly Bencze
PP24966. If ak > 0 (k = 1, 2, ..., n) , then
�cyclic
a61+1
6a42−15a32+20a22−15a2+6≥ n n
�n�
k=1
ak.
Mihaly Bencze
PP24967. If x > 0 and λ ∈ (−∞, 0) ∪ [1,+∞) thenn�
k=1
��x+ k−1
k
� 1λ
�≤�nλ−1 [nx]
� 1λ where [·] denote the integer part.
Mihaly Bencze
PP24968. If ak ≥ 1 (k = 1, 2, ..., n) then� a51+4
a42−2a32+2a23−a3+1≥ 5n.
Mihaly Bencze
PP24969. Let ABC be a triangle. Determine all points M in the plane ofthe triangle such that (
�MA) (
�MA ·MB) ≥�MA ·BC2.
Mihaly Bencze
Proposed Problems 253
PP24970. If a (x, y) = x+y2 , a (x, y) =
√xy and h (x, y) = 2
1x+ 1
y
then solve in
R+ the following system
a (x1, x2) + h (x2, x3) = 2g (x4, x5)a (x2, x3) + h (x3, x4) = 2g (x5, x6)−−−−−−−−−−−−−−−−a (xn, x1) + h (x1, x2) = 2g (x3, x4)
.
Mihaly Bencze
PP24971. Let f : R → R be a convex function. Determine all a, b > 0 anda+ b = 1 such that f (x) + f (y) + f (z) +
�f (ax+ by) ≥ 2
�f (bx+ ay)
for all x, y, z ∈ R.
Mihaly Bencze
PP24972. Let f : R → R be a convex function. Determine all a, b, c > 0and a+ b+ c = 1 such thatf (x)+ f (y)+ f (z)+ f (ax+ by + cz)+ f (ay + bz + cx)+ f (az + bx+ cy) ≥≥ 2f (2bx+ 2cy) + 2f (2by + 2cz) + 2f (2bz + 2cx) for all x, y, z ∈ R.
Mihaly Bencze
PP24973. Solve in R the following system(3x1+x2 + 4x3 + 5x4) (3x2+x3 + 4x4 + 5x5) == (3x2+x3 + 4x4 + 5x5) (3x3+x4 + 4x5 + 5x6) = ...= (3xn+x1 + 4x2 + 5x3) (3x1+x2 + 4x3 + 5x4) = 324.
Mihaly Bencze
PP24974. Determine all functions f : R → R for whichf (x) + 2f (y) + f (z) ≥ xf (y) + yf (z) for all x, y, z ∈ R.
Mihaly Bencze
PP24975. Solve in Z the equation x2 + y3 + z4 + t6 = u5.
Mihaly Bencze
PP24976. Let f : R → R be a convex function. Determine all a, b > 0 anda+ b = 1 such that f (x) + f (y) + f (z) + 3f
�x+y+z3
�≥
≥� f (ax+ by) +�
f (bx+ ay) for all x, y, z ∈ R.
Mihaly Bencze
254 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24977. Solve in R the following system(2x1 + 3x2) (5− x3) = (2x2 + 3x3) (5− x4) = ... = (2xn + 3x1) (5− x2) = 20.
Mihaly Bencze
PP24978. Determine all n, k ∈ N for which (n+ 1)3 + k3 is prime.
Mihaly Bencze
PP24979. If a, b, c > 1 then solve in R the following system:(ax1 + bx2 + cx3) (a+ b+ c+ x4 + x5 − 2x6) == (ax2 + bx3 + cx4) (a+ b+ c+ x5 + x6 − 2x7) = ...= (axn + bx1 + cx2) (a+ b+ c+ x3 + x4 − 2x5) = (a+ b+ c)2 .
Mihaly Bencze
PP24980. If a, b, c > 0 and a+ b+ c = 1, then� a4
1−a3+ 1
3
� (2a+b)4
27−(2a+b)3≥ 2
3
� (a+2b)4
27−(a+2b)3.
Mihaly Bencze
PP24981. If a, b, c > 0 and a+ b+ 1 = 1, then� a4
1−a3+ 1
4
� (a+1)4
64−(a+1)3≥� (1−a)4
4−(1−a)3.
Mihaly Bencze
PP24982. If a, b, c > 0 and a+ b+ c = 1 then 126 +
� a4
1−a3≥ 1
2
� (1−a)4
3−2a−a2.
Mihaly Bencze
PP24983. If a, b, c > 0 and a+ b+ c = 1, then126 +
� a4
1−a3≥ 1
3
� (2a+b)4
27−(2a+b)3+ 1
3
� (a+2b)4
27−(a+2b)3.
Mihaly Bencze
PP24984. Let be A ∈ Mn (R) where n ≥ 2 such that Ak +Ak+1 = On.Determine all k ∈ N for which In −A (A+ In) is invertable.
Mihaly Bencze
Proposed Problems 255
PP24985. Determine all functions f : R → [0,+∞) for which fk isdifferentiable and
�fk�′= f where k ∈ N, k ≥ 2.
Mihaly Bencze
PP24986. Solve in R the equation�6x + 13
1x + 4x+
1x
��13x + 6
1x + 4x+
1x
�= 1225.
Mihaly Bencze
PP24987. If a, b, c > 0 then� a+1
(2a+b+c)2≤ 1
16
� a+1a2
.
Mihaly Bencze
PP24988. Let f : N∗ → N∗ a bijective function. Compute
limn→∞
n�1− f(n)
n
�.
Mihaly Bencze
PP24989. Solve in R the equation4x + 6x + 8x + 2016x = 5x + 7x + 9x + 2013x.
Mihaly Bencze
PP24990. Sove in N the equation 2x+y + (x+ y)2 = 2z+t + (z + t)2 .
Mihaly Bencze
PP24991. Compute
π2�0
x sin 2xdxsin8 x−2 sin6 x+sin4 x−1
.
Mihaly Bencze
PP24992. Compute In =� (2x3+15x2+41x+40)dx
(x2+5x+7)nwhen n ∈ N.
Mihaly Bencze
PP24993. If x1 = 1 and (n+ 1)xn+1 = xn + 1n for all n ≥ 1 then compute
limn→∞
n�1− n2xn
�.
Mihaly Bencze
256 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP24994. If 0 < a1 < a2 < ... < a2n then solve the equation
(ax1 + ax2n) +�ax3 + ax2n−2
�+ ... =
�ax2 + ax2n−1
�+�ax4 + ax2n−3
�+ ...
Mihaly Bencze
PP24995. If Ak ∈ Mn (C) (k = 1, 2, ..., n) such that A1 +A2 + ...+Am−1 == TAm ;A2 +A3 + ...+Am = TA1 ; ...;Am +A1 + ...+Am−2 = TAm−1 thenm�k=1
Ak = On.
Mihaly Bencze
PP24996. If a, b, c ∈ C, |a| = |b| = |c| = 1 and |a+ b+ c| ≤ 1 then��a2 + bc��+��b2 + ca
��+��c2 + ab
�� ≥ |a+ b|+ |b+ c|+ |c+ a| .
Mihaly Bencze
PP24997. If a, b > 1 then solve in R the following system:
abx1 + bax2 = abx2 + b
ax3 = ... = abxn + b
ax1 = ab + ba.
Mihaly Bencze
PP24998. If xk > 0 (k = 1, 2, ..., n) , then
a� x2+x3+...+xn
x21
+ b�
x21
�1x2
+ 1x3
+ ... 1xn
�≥ 2 (n− 1)
√ab for all a, b > 0.
Mihaly Bencze
PP24999. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1, thenn�
k=1
x4k
1−x3k≥ 1
n3−1
(A generalization of problem IX.378, RMT)
Mihaly Bencze
PP25000. If a1 = 2 and (n− 1)xn = (n+ 1) (x1 + x2 + ...+ xn−1) for alln ≥ 2 then prove that
1).n�
k=1
xk21−k = n(n+3)
2 2).∞�n=1
4n−1
x2n
= π2
6 − 1
3).n�
k=1
2k−1
kxk= n
n+1 4).n�
k=1
xkk+1 = 2n − 1
Mihaly Bencze
Proposed Problems 257
PP25001. If x1 = a ∈ R and xn+1 = 4xn (1− xn) for all n ≥ 1 thendetermine all a for which x2016 = 0.
Mihaly Bencze
PP25002. If x1 = 2 and (n− 1)xn = (n+ 1) (x1 + x2 + ...+ xn−1) for alln ≥ 2 then compute
1).∞�n=1
1xn
2).∞�n=1
1xnxn+1
Mihaly Bencze
PP25003. Let ABC be a triangle. Solve in R the following system:�x
(x2+1)ma= y
(y2+1)mb= z
(z2+1)mc
xy + yz + zx = 1.
Mihaly Bencze
PP25004. Determine the number of maximal of nonzero terms of the sum�i,j=1
|f (i)− f (j)| for the all possible f : {1, 2, ..., n} → {a1, a2, ..., ak} when
1 ≤ k ≤ n− 1 and a1, a2, ..., ak ∈ R are given.
Mihaly Bencze
PP25005. Compute limn→∞
n
�2(π+2)
π − 1n2
n�k=1
(n+ k) cos (n+1−k)π2n+1
�
Mihaly Bencze
PP25006. Solve in C the following system
�5x21 + 3x2 + 2
� �5x22 − 3x3 + 2
�= 31x1x2�
5x22 + 3x3 + 2� �
5x23 − 3x4 + 2�= 31x2x3
−−−−−−−−−−−−−−−−−−−�5x2n + 3x1 + 2
� �5x21 − 3x2 + 2
�= 31xnx1
.
Mihaly Bencze
PP25007. Let ABC be a triangle. Determine all points A1, B1, C1 in theplane of the given triangle such thatMA1 +MB1 +MC1 ≤ 1
2 (MA+MB +MC) + 32MG for all M in the plane
of the given triangle.
Mihaly Bencze
258 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25008. If a, b, c > 1 then solve in R the following system:
ax1 + bx2 + cx3 = (a+ b+ c− 3)x4 + 3ax2 + bx3 + cx4 = (a+ b+ c− 3)x5 + 3−−−−−−−−−−−−−−−−−−axn + bx1 + cx2 = (a+ b+ c− 3)x3 + 3
.
Mihaly Bencze
PP25009. If a, b ∈ C are given, then solve the following system(x+ a) y + b = (y + a) z + b = (z + a) t+ b = (t+ a)x+ b = 0.
Mihaly Bencze
PP25010. If xk > 0 (k = 1, 2, ..., n) then
1).�
cyclic
x21
x2≥
n�k=1
xk +1n
��
cyclic
|x1−x2|√x1
�2
2). nn�
k=1
x2k ≥n�
k=1
x2k +2
n(n−1)
��
1≤i<j≤n|xi − xj |
�2
Mihaly Bencze
PP25011. Let ABCDE be a convex pentagon in which DC = DE andBCD∡ = DEA∡ = 90◦. Determine all points F ∈ (AB) such thatFCE∡ = ADE∡ and FEC∡ = BDC∡.
Mihaly Bencze
PP25012. If x0, x1 > 1 and xn+1 = log3 (1 + xn−1 + xn) for all n ≥ 1 thencompute lim
n→∞n (1− xn) .
Mihaly Bencze
PP25013. Let ABC be an acute triangle, AI,BI, CI intersect thecircumcircle in points A1, B1, C1 and AH,BH,CH intersect the circumcirclein points A2, B2, C2. Prove that
�(AI)a (AH)tgA ≤� (A1I)
a (A2H)tgA .
Mihaly Bencze
PP25014. Compute�
1≤i<j≤n
(2i)!(2j)!i!j!(i+j)! .
Mihaly Bencze
Proposed Problems 259
PP25015. Compute limn→∞
�1≤i<j≤n
�i!j!(i+j)!(2i)!(2j)!
�3.
Mihaly Bencze
PP25016. Let ABC be a triangle, the medians AD,BE and CF intersect
the circumcircle in points M,N and P. Prove that (AM+BN+CP )2
AD+BE+CF ≥ 16√3s
9 .
Mihaly Bencze
PP25017. In all acute triangle ABC holds
�� HAHA1
�ctgA≤�
4s2r2
(s2−(2R+r)2)(s2−4Rr−r2)
� s2−4Rr−r2
2sr
when AH,BH,CH
intersect the circumcircle in points A1, B1, C1.
Mihaly Bencze
PP25018. Prove that n! < exp�n(n+1)
2e
�.
Mihaly Bencze
PP25019. Determine all prime pk (k = 1, 2, ..., n+ 1) for which
1p2n+1
=n�
k=1
1p2k.
Mihaly Bencze
PP25020. If λ = limn→∞
n�k=2
�k ln�2k+32k−3
�− 3�, then compute
limn→∞
n
�λ−
n�k=2
�k ln�2k+32k−3
�− 3��
.
Mihaly Bencze
PP25021. Solve in Z the equation�x+ y
2
� �2y + z
3
� �3z + x
4
�= 91
8 .
Mihaly Bencze
PP25022. Determine all prime p for which exist k, n ∈ N such thatnk + k + 1 and (n+ 1)k + k + 1 is divisible by p.
Mihaly Bencze
260 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25023. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1 then
n�k=1
x3k
x2k+xk+1
≥ 1n2+n+1
.
Mihaly Bencze
PP25024. Prove thatn�
k=1
k1k < ne
1e .
Mihaly Bencze
PP25025. Prove thatn�
k=1
eke > n(n+1)
2 .
Mihaly Bencze
PP25026. If a, b, c > 0 then�� a3
a2+1
�2+ 1
16
�� (2a+b+c)3
(2a+b+c)2+16
�2≥ 1
2
�� (a+b)3
(a+b)2+4
�2.
Mihaly Bencze
PP25027. If a, b, c, d are the sides of a convex quadrilateral then�� (a+b+c−d)3
(a+b+c−d)2+1
�2≥�
�(a+b)3
(a+b)2+1
�2.
Mihaly Bencze
PP25028. If a, b, c > 0 and a+ b+ c = 3, then34 +�� a3
a2+1
�2≥ 1
9
�� (2a+b)3
(2a+b)2+9
�2+ 1
9
�� (a+2b)3
(a+2b)2+9
�2.
Mihaly Bencze
PP25029. If a, b, c > 0 then�� a3
a2+1
�2+ 1
9
�� (2a+b)3
(2a+b)2+9
�2≥ 2
9
�� (a+2b)3
(a+2b)2+9
�2.
Mihaly Bencze
PP25030. If a, b, c > 0 and a+ b+ c = 3 then34 +�� a3
a2+1
�2≥ 1
2
�� (a+b)3
(a+b)2+4
�2.
Mihaly Bencze
Proposed Problems 261
PP25031. Solve in N∗ the following equation�1 + 1
x
� �1 + 2
y
� �1 + 3
z
�= 8.
Determine all solutions in Z∗.
Mihaly Bencze and Bela Kovacs
PP25032. If x > 1 thenn�
k=1
1k2
�x+ x2
22+ ...+ xk
k2
�> n
n+1 .
Mihaly Bencze
PP25033. In all triangle ABC holds1). (�
ma)3 ≥�m3
a + 3mambmc
2).��
cos A2
�3 ≥� cos3 A2 + 3s
4R
Mihaly Bencze
PP25034. If a > 0 and a �= 1 thena2n+2−1a2−1
+ a(an−1)a−1 + n+ 1 > an+1−1
a−1 + (n+ 1) a.
Mihaly Bencze
PP25035. Prove that�1 + 6
π
�π �1 + 6
e
�e> 576.
Mihaly Bencze
PP25036. If ak ∈ R (k = 1, 2, ..., n) , n ≥ 3 and a12 < a2
3 < ... < ann+1 then
n�k=1
a1+a2+...+akk2(k+3)ak
< n2(n+1) .
Mihaly Bencze
PP25037. In all triangle ABC holds12
��wa
ma−wa+�
hawa−ha
+�
mama−ha
�≥ wa
ma+ ha
wa+ ma
2ma−haand his
permutations.
Mihaly Bencze
PP25038. Prove thatn�
k=1
(k + 1)��
1 + 1k
�k − 1�< n+ 2n−1
6n + 11n(n+1)12 .
Mihaly Bencze
262 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25039. Prove that 3n−12·3n−1 +
n�k=1
�1 + 1
3k
�3k−1
< 3n+ 32n−18·32n for all n ∈ N∗.
Mihaly Bencze
PP25040. Determine all x, y ∈ R for which
e2√2 ln a ln b ≤ asinx+sin ybcosx+cos y ≤ e2
√ln2 a+ln2 b when a, b > 0.
Mihaly Bencze
PP25041. Prove thatn�
k=1
loga�ak − 1
�loga�ak + 1
�loga�a2k + 1
�< 16n2(n+1)2
27 when a > 1.
Mihaly Bencze
PP25042. If xk > 1 (k = 1, 2, ..., n) , thenn�
k=1
lgn�
n�k=1
xk
�≥
m�k=1
lg xk.
Mihaly Bencze
PP25043. Prove thatn�
k=1
�k√1! + k
√2! + ...+ k
√k!�≤ n(2n2+21n+13)
6 .
Mihaly Bencze
PP25044. Compute λ = limn→∞
1
(2nn )
�n�
k=0
(nk)2
n−k+1
�2
and
limn→∞
n
�λ− 1
(2nn )
�n�
k=0
(nk)n−k+1
�2�.
Mihaly Bencze
PP25045. Prove thatn�
k=1
�3√1 · 2 + 3
√2 · 3 + ...+ 3
�k (k + 1)
�≤ n(n+1)(n+5)
9 .
Mihaly Bencze
PP25046. Compute limn→∞
n
�γ − ln 2−
�n�
k=1
(−1)k−1
k
�nk
�− ln 2k
��when
γ = 0, 92 is the Euler constant.
Mihaly Bencze
Proposed Problems 263
PP25047. Solve in R the following system
5− x =�5− y2
�2
5− y =�5− z2
�2
5− z =�5− x2
�2.
Mihaly Bencze
PP25048. Prove that√1 +
√2 + ...+
√n2 − 1 ≥ 2n(n2−1)
3 .
Mihaly Bencze
PP25049. Compute limn→∞
�n2�arctg(n+1)
n+1 − arctgnn
�− π
2
�.
Mihaly Bencze
PP25050. Compute limn→∞
n
�ln 2−
n�k=1
sinh 1n+k
�.
Mihaly Bencze
PP25051. Compute limn→∞
n
�1−
�
1≤i<j≤n
1ij
�
1≤i<j≤n
1(i+1)(j+1)
�.
Mihaly Bencze
PP25052. Solve in M3 (C)×M3 (C) the following system:�X3 + Y 3 = O3
X−1 + Y −1 = I3.
Mihaly Bencze
PP25053. Compute λ = limn→∞
argzn where zn =�i+ 12
� �i+ 22
�...�i+ n2
�
and limn→∞
n (λ− arg zn) .
Mihaly Bencze
PP25054. Determine all a > 0 for whicha2 ≤ (a− 1)
�a+1a−1
�x+ (a+ 1)
�a−1a+1
�x≤ a3 for all x ∈ [0, 1] .
Mihaly Bencze
264 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25055. Prove thatn�
k=1
�k
�k�k√k ≤ n(15n+17)
2 .
Mihaly Bencze
PP25056. If A,B,C ∈ Mn (z) , n ≥ 3 are nonsingular matrices such that(A∗B∗)∗ = BA, (B∗C∗)∗ = CB, (C∗A∗)∗ = AC, then
detA+ detB + detC <√52 .
Mihaly Bencze
PP25057. Let A ∈ Mn (R) such that TrA = Tr (A∗) = 0. Determine alln ∈ N for which det (An + nIn) det (A
n + In) ≥ 0.
Mihaly Bencze
PP25058. Solve in R the following system
1 + 6x2+y2+z2
= 1x + 1
y + 1z
1 + 6y2+z2+t2
= 1y + 1
z + 1t
1 + 6z2+t2+x2 = 1
z + 1t +
1x
1 + 6t2+x2+y2
= 1t +
1x + 1
y
.
Mihaly Bencze
PP25059. Solve in Z the equation 1 + 6x2+y2+z2
= 1x + 1
y + 1z .
Mihaly Bencze
PP25060. Solve in Z the equation 1 + n(n+1)n�
k=1
x2k
=n�
k=1
1xk.
Mihaly Bencze
PP25061. Prove that�
1≤i<j≤n
3
�(n+1)2
n2ij(i+1)(j+1)≤ (n−1)(n+2)
6 .
Mihaly Bencze
PP25062. In all triangle ABC holds� 1
w2λa
≥ 3�
2s2−r2−4Rr
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
Proposed Problems 265
PP25063. In all triangle ABC holds�� a
b+c−a ≥ 4(s2−r2−Rr)s2+r2+2Rr
.
Mihaly Bencze
PP25064. In all triangle ABC holds� ab2
a+b ≤ s2 − r2 − 4Rr.
Mihaly Bencze
PP25065. Determine all n ∈ N for which
�n�
k=1
1|sin k|
�≥ 2016 when [· ]
denote the integer part.
Mihaly Bencze
PP25066. Determine all n ∈ N for which
�n�
k=1
1|cos k|
�≥ 2016 when [· ]
denote the integer part.
Mihaly Bencze
PP25067. Determine all n ∈ N for which
�n�
k=1
tg 1k
�≥ 2016 when [· ] denote
the integer part.
Mihaly Bencze
PP25068. Determine all n ∈ N for which
�n�
k=1
ctg 1k
�≥ 2016 when [· ]
denote the integer part.
Mihaly Bencze
PP25069. Determine all n ∈ N∗ for which ne < (n+1)n+1
nn < (n+ 1) e.
Mihaly Bencze
PP25070. Denote pn the nth prime and xn = p1p1−1 · p2
p2−1 ...pn
pn−1 . Determineall n ∈ N for which xn < 2016.
Mihaly Bencze
266 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25071. If xk ∈ [0, 1] (k = 1, 2, ..., n) thenn�
k=1
�xk (1− xk) +
n�k=1
xk ≤ (1+√10)n
2 .
Mihaly Bencze
PP25072. In all triangle ABC holds� 1√
2ab+√3bc+2ca
≥ 98(s2−r2−4Rr)
.
Mihaly Bencze
PP25073. If ak > 0 (k = 1, 2, ..., n) and a12 < a2
3 < ... < ann+1 then
n�k=1
(a1 + a2 + ...+ ak) >n!(n+3)!an13·22n+1 .
Mihaly Bencze
PP25074. In all triangle ABC holds��a2mbmc
bcma+ bcma
mbmc− a�≥ 4Rsr.
Mihaly Bencze
PP25075. If a, b, c > 0 then��a2+b2
a+b
�3≥ 3abc.
Mihaly Bencze
PP25076. If a, b, c > 1, then��a2
bc + bc− a�3
≥ 3abc.
Mihaly Bencze
PP25077. Solve in Z the equation�x2
yz + yz − x��
y2
zx + zx− y��
z2
xy + xy − z�= xyz.
Mihaly Bencze
PP25078. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1 then�
1≤i<j≤nxixj ≤ n−1
2n .
Mihaly Bencze
PP25079. If xj > 0 (j = 1, 2, ..., n) andn�
k=1
xk = 1 compute
max�
1≤i1<...<ik≤nxi1xi2 ...xik .
Mihaly Bencze
Proposed Problems 267
PP25080. Prove thatn�
k=1
k8+k5+1k3+1
≥ n(n+1)2 .
Mihaly Bencze
PP25081. If xk ∈�0, π2�(k = 1, 2, ..., n) , then
�cyclic
11−sinx1 sinx2
≤n�
k=1
1cos2 xk
.
Mihaly Bencze
PP25082. In all triangle ABC holds�� r2a
a
�λ≥ 33λ+1
�r2
2s
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP25083. In all triangle ABC holds�� a2b
a3+b
�λ≤ sλ
3λ−1 for all λ ∈ [0, 1] .
Mihaly Bencze
PP25084. In all triangle ABC holds��2a2 + b2 + c2
�≥ 4s2
�s2 + r2 + 2Rr
�2.
Mihaly Bencze
PP25085. Prove that 119 +
π1�0
x2tgxdx ≤ 3 ln 22 − 3π2
32 +
π2�0
tg (sinx) dx.
Mihaly Bencze
PP25086. If n ∈ N∗ and f : [0, 1] → R is a continuous function then∞�k=0
�1�0
x2k(k+1)+nf (x) dx
�2
≤ π2
8
1�0
x2nf2 (x) dx.
Mihaly Bencze
PP25087. In all acute triangle ABC holds�
AtgA ≥ 2sπr3(s2−(2R+r)2)
.
Mihaly Bencze
PP25088. In all triangle ABC holds�
AtgA2 ≥ π(4R+r)
3s .
Mihaly Bencze
268 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25089. In all triangle ABC holds�
Ama�
ma+
�
Aha�
ha+
�
Awa�
wa≤ π.
Mihaly Bencze
PP25090. Determine all continuous functions f : R∗ → R∗ for which2�1
f (x) dx+ e52�1
dxf(x) ≤ e4 + e.
Mihaly Bencze
PP25091. Compute limn→∞
n
�3 ln 3− π√
3− 2
n�k=1
1k(3k−2)
�.
Mihaly Bencze
PP25092. Solve in C the following system 1x + 2
y−1 + 3z−2 + 4
t−3 =
= 1y + 2
z−1 + 3t−2 + 4
x−3 = 1z + 2
t−1 + 3x−2 + 4
y−3 = 1t +
2x−1 + 3
y−2 + 4z−3 = 1.
Mihaly Bencze
PP25093. In all triangle ABC holds�
�
�
tgA2+�
tgC2
�
�
tgB2
1+3tgB2
�
tgA2tgC
2
≥√3.
Mihaly Bencze
PP25094. In all triangle ABC holds 272 ≤� ctgA
2ctgB
2
1−tgA2tgB
2
≤ s2
2r2.
Mihaly Bencze
PP25095. If xk ≥ 0 (k = 1, 2, ..., n) andn�
k=1
xk = n then
n�k=1
exk+2e2xk+exk+1
≥ n(e+2)e2+e+1
.
Mihaly Bencze
PP25096. If a, b ∈ [0, 1] then
a+2a2+a+1
+ b+2b2+b+1
≥ 1 + ab+2(ab)2+ab+1
+(1−a)(1−b)(1−ab)(
√a+2
√b)(
√b+2
√a)
√ab
(a2+a+1)(b2+b+1)((ab)2+ab+1).
Mihaly Bencze
Proposed Problems 269
PP25097. If ak > 0 (k = 1, 2, ..., n) then�
cyclic
a1an−11 +(n−1)a2a3...an
≤
�
n�
k=1ak
�2
n2n�
k=1ak
.
Mihaly Bencze
PP25098. In all triangle ABC holds� 1
sin2 A cos2 A2
≥ 163 .
Mihaly Bencze
PP25099. If ak > 0 (k = 1, 2, ..., n) then� a1an−12 (a2+...+an)
n ≥ nn
(n−1)n(�
a1a2)n−1 .
Mihaly Bencze
PP25100. If ak > 0 (k = 1, 2, ..., n) then��
cyclic
a1a2
�2
≥ 2
�n�
k=1
ak
��n�
k=1
1ak
�− n2.
Mihaly Bencze
PP25101. If 1 ≥ b ≥ 0 and a1 = 0 and 2an+1 = b+ a2n for all n ≥ 1 thencompute lim
n→∞n�1−
√1− b− xn
�.
Mihaly Bencze
PP25102. If ak > 1 (k = 1, 2, ..., n) then�
ax1 loga1a2a3...an =
�n�
k=1
ak
�x
.
Mihaly Bencze
PP25103. If xk > 0 (k = 1, 2, ..., n) and
fa (x1, x2, ..., xn) =(xa
1+xa2+...+xa
k)(xa2+xa
3+...+xak+1)...(x
an+xa
1+...+xak−1)
n�
k=1aak
when
k ∈ {1, 2, ..., n} then fa (x1, x2, ..., xn) ≥ fb (x1, x2, ..., xn) for all a ≥ b > 0.
Mihaly Bencze
PP25104. If ak > 0 (k = 1, 2, ..., n) then� an1
a2a3...an≥
n�k=1
ak.
Mihaly Bencze
270 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25105. If ap > 0 (p = 1, 2, ..., n) and k ∈ {1, 2, ..., n− 1} , then�
cyclic
(a1a2...ak)n+1
(a1a2...an)k ≥� a1...ak.
Mihaly Bencze
PP25106. In all triangle ABC holds2(s2−r2−Rr)s2+r2+2Rr
+ 3
�2Rr
s2+r2+2Rr≥ 2.
Mihaly Bencze
PP25107. In all triangle ABC holds s2+r2
2Rr + 3�
rR ≥ 8.
Mihaly Bencze
PP25108. Solve in Z the equation x3
y+z2+ y3
z+x2 + z3
x+y2= 3a2
1+a when
a ∈ Z ∈ \ {−1} is given.
Mihaly Bencze
PP25109. In all acute triangle ABC holdsR(s2+8Rr+7r2)−2r(s2−r2)
r(s2+2Rr+r2)+ 3
�R(s2−(2R+r)2)
r(s√2+r
√2+2Rr)
≥ 2.
Mihaly Bencze
PP25110. If x, y, z > 1 then�
logxy z +3
��logxy z ≥ 2.
Mihaly Bencze
PP25111. If x, y, z > 0 then� x2+yz
x(y+z) + 2 3
�xyz
(x+y)(y+z)(z+x) ≥ 4.
Mihaly Bencze
PP25112. If xk > 0 (k = 1, 2, ..., n) then
�cyclic
x1x2+...+xn
+n
����n�
k=1xk
�
cyclic(x1+...+xn−1)
≥ n+1n−1 .
Mihaly Bencze
PP25113. If a ≥ 1 then 2�a9 + 2a
�≥ 3a3 + 2
�a6 − 1
�√a6 − 1.
Mihaly Bencze
Proposed Problems 271
PP25114. Solve in R the following system
x+ 12 3√y ≥ 3
2 + (z − 1)√t− 1
y + 12 3√z
≥ 32 + (t− 1)
√x− 1
z + 12 3√t
≥ 32 + (x− 1)
√y − 1
t+ 1
2√
3√x≥ 3
2 + (y − 1)√z − 1
.
Mihaly Bencze
PP25115. Prove thatn�
k=2
√k−1
2k52−3k
32+k
76≤ n−1
2n
Mihaly Bencze
PP25116. If ak > 1 (k = 1, 2, ..., n) then
4n�
k=1
1x2k+
n�k=1
�2x8
k−3x2k+4
x6k−1
�2≥ 4
n3
�n�
k=1
ak
�4
.
Mihaly Bencze
PP25117. Prove that 4n+n�
k=2
�2k3−3k+4k
13
k2−1
�2
≥ 2n(n+1)(2n+1)3 .
Mihaly Bencze
PP25118. If x, y, z > 0 then(x+ y) (y + z) (z + x) ≥ 8xyz + 1
3 (�√
x |y − z|)2 .
Mihaly Bencze
PP25119. If x, y, z > 0 then�� x
y+z
�2+ 14xyz
(x+y)(y+z)(z+x) ≥ 4.
Mihaly Bencze
PP25120. If ak > 1 (k = 1, 2, ..., n) and λ ≥ 1 thenn�
k=1
ak +1
n�
k=1
aλk
≥n�
k=1
1aλ−1k
.
Mihaly Bencze
272 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25121. If a ∈�0, π2�then
3
(1+sin2 a)(1+cos2 a)+ 2 sin (sin a+ cos a) cos (sin a− cos a) < 1
sin2 a cos2 a.
Mihaly Bencze
PP25122. Determine all x, y, z ∈ Z for which
sin 2x < 1y2(2−z2)
sin 2y < 1z2(2−x2)
sin 2z < 1x2(2−y2)
.
Mihaly Bencze
PP25123. If xk > 0 (k = 1, 2, ..., n) , thenn�
k=1
x2n−3k ≥ 1
2n−2 |(x1 − x2) (x2 − x3) ... (xn − x1)|+
+ n−1
���xn−11 x2...xn−1
� ��x1x
n−12 x3...xn−1
�...��
x1x2...xn−2xn−1n−1
�.
Mihaly Bencze
PP25124. In all triangle ABC holds� r3a
(tgA2+tgC
2 )2tg2 B
2
≥ (4R+r)3−12s2R36 .
Mihaly Bencze
PP25125. In all triangle ABC holds� ra
2rb+rc+ s2
3((4R+r)2−2s2)≥ 4
3 .
Mihaly Bencze
PP25126. In all triangle ABC holds� −a+b+c
3a−b+c + r(4R+r)3(s2−2r2−8Rr)
≥ 43 .
Mihaly Bencze
PP25127. If ak ∈ (0, 1] (k = 1, 2, ..., n) then determine all λ ∈ R for which
n�k=1
(1−ak)λ+1
aλk≤
�
1−n�
k=1ak
�λ+1
n�
k=1aλk
.
Mihaly Bencze
PP25128. If a, b > 0 then�a5 + 4a2b3 + b6
� �b5 + 4b2a3 + a6
�≥�
ab�a3 + 2b3
�+ b2
�2a3 + b3
�� �ab�b3 + 2a3
�+ a2
�2b3 + a3
��.
Mihaly Bencze
Proposed Problems 273
PP25129. If x ∈�0, π2�then
sinn x1+sinx+...+sinn x + cosn x
1+cosx+...+cosn x + (n−1)(sin x+cosx)(n+1) sinx cosx ≥ 2n
n+1 .
Mihaly Bencze
PP25130. Solve in R the following system
x21 + 2 {x2}2 = 5 [x3]2
x22 + 2 {x3}2 = 5 [x4]2
−−−−−−−−−x2n + 2 {x1}2 = 5 [x2]
2
where
[·] and {·} denote the integer, respective the fractional part.
Mihaly Bencze
PP25131. Compute limn→∞
1�0
1�0
ln (xn + ey) ln (yn + ex) dxdy.
Mihaly Bencze
PP25132. Determine all a, b, c ∈ N for which xm,n = (am)!(bn)!m!n!(m+n)! and
xm,n−1 + xm−1,n = cxm−1,n−1 for all m,n ∈ N∗.
Mihaly Bencze
PP25133. If n ∈ N , n ≥ 2 and 1 = d1 < d2 < ... < dk = n are the divisorsof n, then determine all r ∈ {1, 2, ..., n} for which n is prime if and only ifd1d2...dr + d2d3...dr+1 + ...+ dnd1...dr−1 = nk (r − 1) .
Mihaly Bencze
PP25134. In all scalene triangle ABC holds� m2a
(ma−mb)2(ma−mc)
2 ≥ 23(s2−r2−4Rr)
Mihaly Bencze
PP25135. Determine all a, b, c ∈ Z for which the equation�a2 + b2
�x2 − 2
�b2 + c2
�x−�c2 + a2
�= 0 have rational roots.
Mihaly Bencze
PP25136. In all triangle ABC holds��
tgA2 tg
B2 + 3
√3�
ctgB2
�ctgA
2 ctgC2 ≥ 28
√3.
Mihaly Bencze
274 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25137. Solve in Z the equation x1x2(1+x3)
+ x2x3(1+x4)
+ ...+ xnx1(1+x2)
= nk+1
when k ∈ Z\ {−1} is given.
Mihaly Bencze
PP25138. If λ = limn→∞
√n (2n−1)!!
(2n)!! then compute limn→∞
n�λ−√
n · (2n−1)!!(2n)!!
�.
Mihaly Bencze
PP25139. If xk ∈�0, π2�(k = 1, 2, ..., n) then
n�k=1
11−(sinxk)
n−1 ≥ �cyclic
11−sinx1 sinx2... sinxn−1
.
Mihaly Bencze
PP25140. Prove thatn�
k=1
����1
3√
k(k+1)− sin 1
3√
k(k+1)
���� <6nn+1 .
Mihaly Bencze
PP25141. In all triangle ABC holds�� b+c
2
�2cos2 A
2 ≤ 6�s2 − r2 − 4Rr
�.
Mihaly Bencze
PP25142. If xk > 0 (k = 1, 2, ..., n) then�
cyclic
�x1 −
√x1x2 + x2
�2 ≥n�
k=1
x2k.
Mihaly Bencze
PP25143. If xk > 0 (k = 1, 2, ..., n) then determine all p ∈ N for whichn�
k=1
xk ≥ p n
�1n
n�k=1
xnk + (n− p) n
�n�
k=1
xk.
Mihaly Bencze
PP25144. If λ > 0 and zk ∈ C∗ (k = 1, 2, ..., n) , thenn�
k=1
���zk + λzk
���4≥ 4nλ2 + 8λ
n�k=1
Re�z2k�.
Mihaly Bencze
Proposed Problems 275
PP25145. Determine all a, b, n,m ∈ N for which(an)! + (bm)! ≤ (n!)a+1 + (m!)b+1 .
Mihaly Bencze
PP25146. Determine all m,nk ∈ N (k = 1, 2, ..., n) for whichm�k=1
1((nk)!)
3 ≤ m�
n�
k=1nk
�
!.
Mihaly Bencze
PP25147. Let AkBkCk (k = 1, 2, ..., n) be triangles. Prove that
�wa1wa2 ...wan ≥ 3
�3�
cyclic
sa1ra1 .
Mihaly Bencze
PP25148. If ak ∈ [0, 1] (k = 1, 2, ..., n) then
�cyclic
√a1a2 +
�cyclic
�a1+a21+a1a2
≤ 2n�
k=1
√ak.
Mihaly Bencze
PP25149. In all triangle ABC holds� ra√
r2a+2(rb+rc)2≥ 1.
Mihaly Bencze
PP25150. If xk > 0 (k = 1, 2, ..., n) , then��{x1}2
x2+ [x1]
2
x3
�≥
n�
k=1x2k
n�
k=1xk
when
[·] and {·} denote the integer, respective the fractional part.
Mihaly Bencze
PP25151. Determine all polynomials P for which the zeros of P are inarithmetical progression, and the zeroes of P ′ are too in arithmeticalprogression.
Mihaly Bencze
276 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25152. Determine all polynomials P for which the zeros of P are ingeometrical progression, and the zeroes of P ′ are too in geometricalprogression.
Mihaly Bencze
PP25153. Determine all postve integers k, n for which the numberk (k − 1) ...10k (k − 1) ...10...k (k − 1) ...1 is prime.
Mihaly Bencze
PP25154. Let x be a nonzero real number sarisfying the equationa0x
n + a1xn−1 + ...+ an−1x+ an = 0, where a0, a1, ..., an ∈ Z such that
|a0|+ |a1|+ ...+ |an| > 1. Determine all n ∈ N∗ for which|x| > 1
|a0|+|a1|+...+|an| .
Mihaly Bencze
PP25155. Let ABC be a triangle. Determine all λ ∈ R for which�a2�
λra
− rrbrc
�= 4 (R+ (λ+ 1) r) .
Mihaly Bencze
PP25156. Consider n distinct points in the plane, n ≥ 3 arranged such thatthe number r (n) of segments of length d is maximed. Prove that
π2
2 − 154 <
∞�n=3
1r(n) .
Mihaly Bencze
PP25157. In all triangle ABC holds� 1
rstgB
2+2tgA
2tgC
2+2(tgA
2tgC
2 )2 ≥ s2
r(4R+r) .
Mihaly Bencze
PP25158. Solve in Z the following system:
�x+ y2
� �y2 + z
�= (z + x)3�
y + z2� �
z2 + x�= (x+ y)3�
z + x2� �
x2 + y�= (y + z)3
.
Mihaly Bencze
Proposed Problems 277
PP25159. Solve in C the following system
x4 − y + 1 = z2
y4 − z + 1 = x2
z4 − x+ 1 = y2.
Mihaly Bencze
PP25160. Determine all xk ∈ N∗ (k = 1, 2, ..., n) for which nx1x2...xn−1−11+x1x2...xn
isinteger.
Mihaly Bencze
PP25161. If a, b, c > 0 then� 1√
a3+2b3+6abc≤ 1√
abc.
Mihaly Bencze
PP25162. Denote pn the nth prime number. If n ≥ k (k + 1) then pn > knfor all k ∈ N∗, k ≥ 2.
Mihaly Bencze
PP25163. If a, b, c > 0 and� a2+1
a = 4 then
a√bc+ b
�c�1a + 1
b +1c
�+ c�a�1a + 1
b +1c
�+�1a + 1
b +1c
�√ab ≤
≤ 2�1 +�abc�1a + 1
b +1c
��.
Mihaly Bencze
PP25164. If xk ∈ R thenn�
k=1
|cosxk|+�
cyclic
|cos (x1 − x2)| ≥ n− 1.
Mihaly Bencze
PP25165. Compute limn→∞
n
�14 − ln
�12n
n�k=1
�2 + k
n2
���.
Mihaly Bencze
PP25166. In all triangle ABC holds� sin A
2
(cos A2 )
2 ≥ 2.
Mihaly Bencze
278 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25167. If b1, b2, ..., bn is a rearangement of numbers a1, a2, ..., an > 0 then
�1≤i<j≤n
aibjai+bj
≤ n
2n�
k=1ak
��
1≤i<j≤naibj
�.
Mihaly Bencze
PP25168. Let ABC be a triangle. Determine all λ > 0 for which� (a+b−c)λ
aλ+bλ−cλ≤ 3.
Mihaly Bencze
PP25169. Solve in C the following system
x21 +�x22 + 32 =
�x23 + 96
x22 +�x23 + 32 =
�x24 + 96
−−−−−−−−−−−−x2n +
�x21 + 32 =
�x22 + 96
Mihaly Bencze
PP25170. In all triangle ABC holds��a(b+c)
a2+bc≤�
2(5s2+r2+4Rr)s2+r2+2Rr
.
Mihaly Bencze
PP25171. Let n ≥ 2 be an integer. Find the number of integers k with0 ≤ k < n and such that k3 leaves a remainder of 1 when divided by n.
Mihaly Bencze
PP25172. Solve in R the following system:
�x2 − 2y + 6 log3 (9− 2z) = x�y2 − 2z + 6 log3 (9− 2x) = y√z2 − 2x+ 6 log3 (9− 2y) = z
.
Mihaly Bencze
PP25173. If λ ∈ (0, 1) then determine all x ∈ R for which
�(a+ x) (b+ x) = x+
�aλ+bλ
2
� 1λ.
Mihaly Bencze
Proposed Problems 279
PP25174. If an+1 = [an] {an} for all n ≥ 0 when [·] and {·} denote theinteger part respectively the fractional part. Determine all a0 ∈ R for whichthe sequence (an)n≥0 is periodical.
Mihaly Bencze
PP25175. If M is a set for which cardM = 2n, then the number ofpartitions of M formed by 2-subsets of M is (2n)!
2n·n! .
Mihaly Bencze
PP25176. Let ABC be a triangle. Determine the best constant λ > 0 for
which�
R2r ≥ λ3a2b2c2
�
(λa2−(b−c)2).
Mihaly Bencze
PP25177. Prove that�2k+1
2k
�−�
2k
2k−1
�is divisible by 8k for all k ∈ N.
Mihaly Bencze
PP25178. Determine all functions f : (0,+∞) → (0,+∞) for which(f (x) + f (y) + 2xf (xy)) f (x+ y) = f (xy) for all x, y > 0.
Mihaly Bencze
PP25179. Solve in R the equationn�
k=1
�k3
x
�= n2(n+1)2
4 when [·] denote the
integer part.
Mihaly Bencze
PP25180. If x ∈�0, π2�then�
n�k=1
1+(sinx)4kn
1+(sinx)8kn
��n�
k=1
1+(cosx)4kn
1+(cosx)8kn
�< 1
sin2 x cos2 x.
Mihaly Bencze
PP25181. Solve in R the following equationn�
k=1
�kx
�= n(n+1)
2 when [·]denote the integer part.
Mihaly Bencze
280 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25182. Solve in R the following equationn�
k=1
�k2
x
�= n(n+1)(2n+1)
6 when
[·] denote the integer part.
Mihaly Bencze
PP25183. Compute λ = limn→∞
�n�
k=1
arctgkk2+1
− 12arctg
2n
�and
limn→∞
n
�λ+ 1
2arctg2n−
n�k=1
arctgkk2+1
�.
Mihaly Bencze and Lajos Longaver
PP25184. In all triangle ABC holds� √
sinA+√sinB
�
cos A−B2
cos C2+�
|sin A−B2 | sin C
2
≤ 6.
Mihaly Bencze
PP25185. In all triangle ABC holds−195
8 ≤ 3�
cos 2A− 8�
sinA cosB + 5(2R−s+2r)R ≤ 60.
Mihaly Bencze and Lajos Longaver
PP25186. Ifn�
k=1
|shkx| ≥ n(n+1)2 |shx| .
Mihaly Bencze
PP25187. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1 thenn�
k=1
xarctgxkk ≥ 1.
Mihaly Bencze
PP25188. In all acute triangle ABC holds��√tgA+ 4ctgB +
√tgB + 4ctgA+
√tgA+ 9ctgB +
√tgB + 9ctgA
�≥
≥ 21√2.
Mihaly Bencze
PP25189. In all acute triangle ABC holds
2�
tg2A2 tg2B2 ≥ 1 +
�tg2B2
�12
�tg4A2 + tg4C2
�.
Mihaly Bencze
Proposed Problems 281
PP25190. Solve the following system
x41 + 35x22 + 24 = 10x3�x24 + 5
�
x42 + 35x23 + 24 = 10x4�x25 + 5
�
−−−−−−−−−−−−−−x4n + 35x21 + 24 = 10x2
�x23 + 5
�.
Mihaly Bencze
PP25191. If x ∈ R then 2ex
e2x+1+ e−x
�e4x+1
2 ≤ (ex+1)2
2ex +((x−1)e2x+x+1)
2
2x2ex(e2x+1).
Mihaly Bencze
PP25192. Prove thatn�
k=1
k
(k+2)�
k√1·2+ k√2·3+...+ k√
k(k+1)� ≥ n
n+3 .
Mihaly Bencze
PP25193. In all acute triangle ABC holds
1).� 1
1−sinA sinB ≤�
s2+r2−4R2
s2−(2R+r)2
�2− 8R(R+r)
s2−(2R+r)2
2).� 1
1−cosA cosB ≤�s2+r2+Rr
2sr
�2− 4R
r
Mihaly Bencze
PP25194. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = 1 then
n�k=1
�1 + 1
ak
�≥ (n+ 1)n .
Mihaly Bencze
PP25195. Determine all a, b, c > 0 for which (a+ b+ c)�1a + 1
b +1c
�≤ 81
8 .
Mihaly Bencze
PP25196. Let ABCD be a convex quadrilateral. Determine min and maxof the following expression
�sin A
2 cos B2 tg
C4 .
Mihaly Bencze
PP25197. If ak, bk, ck ∈ N\ {0, 1} thenn�
k=1
�1− 1
a2k
�k �1− 1
b2k
�k �1− 1
c2k
�k> 8n−1
7·8n .
Mihaly Bencze
282 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25198. In all acute triangle ABC holds
R+rR <
� 1A2
�2A�A
sinxx dx
��2A�A
xdxsinx
�< s2+r2−4R2
2(s2−(2R+r)2).
Mihaly Bencze
PP25199. Solve in N the equation xy + yx = (x− y)2(x+y) .
Mihaly Bencze
PP25200. Solve in R the following system
�1 + 3
1x
�(13− 7y) = 16 (2 + 7z)�
1 + 31y
�(13− 7z) = 16 (2 + 7x)�
1 + 31x
�(13− 7x) = 16 (2 + 7y)
.
Mihaly Bencze and Gyorgy Szollosy
PP25201. Solve in N the equationn�
k=1
(1 + xk) = ynn�
k=1
xk.
Mihaly Bencze
PP25202. Solve in R the following system
2x2 + 3 ≥ 3y + 2√z2 − z + 1
2y2 + 3 ≥ 3z + 2√x2 − x+ 1
2z2 + 3 ≥ 3x+ 2�y2 − y + 1
.
Mihaly Bencze
PP25203. If zp ∈ C∗ (p = 1, 2, ..., n) such that |z1| = |z2| = ... = |zn| anda1 =
z1+z2z1−z2
, a2 =z2+z3z2−z3
, ..., an = zn+z1zn−z1
then�a1a2...a2k +
�a1a2...a2k−2 + ...+ 1 = 0 if n = 2k + 1 and�
a1a2...a2k−1 +�
a1a2...a2k−3 + ...+�
a1 = 0 if n = 2k.
Mihaly Bencze
PP25204. If a > 1 and xk > 0 (k = 1, 2, ..., n) then
n
�n�
k=1
loga xk ≤ loga
�1n
n�k=1
xk
�.
Mihaly Bencze
Proposed Problems 283
PP25205. Solve in Z the equationn�
k=1
xk−1xk+1 = n(m−1)
m+1 when m ∈ Z\ {−1} is
given.
Mihaly Bencze
PP25206. Compute∞�n=1
1
[ 3√n3+2n+1]2 when [·] denote the integer part.
Mihaly Bencze
PP25207. If x0 = x1 = 1 and n (n+ 1)�xn+1
√xn−1 − xn
√xn�=
√xn−1xn
for all n ≥ 1 thenn�
k=1
√xk
k(2k−1) ≤n
n+1 .
Mihaly Bencze
PP25208. In all triangle ABC holds�
c�1 + sin A
2
�2 �1 + sin B
2
�2 ≤≤ 3(s2−3r2−12Rr)(3s2+3r2−4Rr)+s2(−9s2+36r2+144Rr)
4sRr ≤≤ 3(s2−3r2−12Rr)(3s2+3r2−4Rr)+9s2(−s2+4r2+16Rr)
4sRr .
Mihaly Bencze
PP25209. If x0 = x1 = 1 and n (n+ 1)�xn+1
√xn+1 − xn
√xn�=
√xn−1xn
for all n ≥ 1 thenn�
k=1
�(k + 1)
√xk�3 ≤ n2
�2n2 − 1
�.
Mihaly Bencze
PP25210. If a, b, c, d > 0 then���
1 + 1a
�a�2 ≥ 3��
1 + 1a
�2a.
Mihaly Bencze
PP25211. If a, b > 0 thenn�
k=1
�a
1k b
k−1k + a
k−1k b
1k
�≤ n (a+ b) .
Mihaly Bencze
PP25212. If a, b, c > 0 then��a+b
2
�n ≥��√
ab�n
+ 13·2n�����
√an −
√bn����2
for all n ∈ N.
Mihaly Bencze
284 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25213. If ak > 0 (k = 1, 2, ..., n) then�
cyclic
�a21 + a22 ≤
√2
n�k=1
ak +√24
�cyclic
(a1−a2)2
a1+a2.
Mihaly Bencze
PP25214. If ak > 0 (k = 1, 2, ..., n) , then determine all n ∈ N for which�
cyclic(a1−a2)
2
2n+1 max{a1,a2,...,an} ≤n�
k=1
ak −�
cyclic
√a1a2 ≤
�
cyclic(a1−a2)
2
2n+1 min{a1,a2,...,an} .
Mihaly Bencze
PP25215. If a, b, c > 0 then determine all λ1,λ2,λ3 > 0 for whicha2
λ1a+λ2b+λ3c+ b2
λ1b+λ2c+λ3a+ c2
λ1c+λ2a+λ3b≥ a+b+c
λ1+λ2+λ3.
Mihaly Bencze
PP25216. If 1 ≤ xk ≤ 2 (k = 1, 2, ..., n) , thenn2(n+1)2
8 ≤n�
k=1
(x1 + x2 + ...+ xk)�
1x1
+ 1x2
+ ...+ 1xk
�2≤ n2(n+1)2
4 .
Mihaly Bencze
PP25217. If a, b, c > 0 then determine all λ ∈ R for which� a
a+λb+c ≥ λ+1λ+2 .
Mihaly Bencze
PP25218. If x, y, z > 0 and x+ y + z = 1 then� x1−x4 + 81
80 ≥ 16� 1−x
16−(1−x)4.
Mihaly Bencze
PP25219. If x, y, z > 0 and x+ y + z = 1, then� x1−x4 + 81
80 ≥ 27� 2x+y
81−(2x+y)4+ 27
� x+2y
81−(x+2y)4.
Mihaly Bencze
PP25220. If x, y, z > 0 and x+ y + z = 1, then� x1−x4 + 27
� 2x+y
81−(2x+y)4≥ 54
� x+2y
81−(x+2y)4.
Mihaly Bencze
Proposed Problems 285
PP25221. If x, y, z > 0 and x+ y + z = 1, then� x1−x4 + 64
� 1+x256−(1+x)4
≥ 16� x+y
16−(x+y)4.
Mihaly Bencze
PP25222. Solve in R the following system:� √x4 + 1 +
�y4 + 1 +
√z4 + 1 = 3
√2
xy + yz + zx = 1
Mihaly Bencze
PP25223. If a, b, c ∈ C then
3�|a|2 + |b|2 + |c|2 + |a+ b+ c|2
�≥ (|a+ b|+ |b+ c|+ |c+ a|)2 .
Mihaly Bencze
PP25224. If a, b, c ∈ C then
4�|a+ b|2 + |b+ c|2 + |c+ a|2
�≥ (|a|+ |b|+ |c|+ |a+ b+ c|)2 .
Mihaly Bencze
PP25225. If ak ∈ [0, 1] (k = 1, 2, ..., n) then�
cyclic
a11+a2+a3+...+an
+n�
k=1
(1− ak) ≤ 1.
Mihaly Bencze
PP25226. If x, y, z > 0 and λ ≥ 1 then��
1 + x1λ
�λ+ 3
�1 +�x+y+z
3
� 1λ
�λ
≥ 2��
1 +�x+y
2
� 1λ
�λ
.
Mihaly Bencze
PP25227. If ak > 1 (k = 1, 2, ..., n) (n ≥ 3) then solve in R the following
system:
n�k=1
axk =
�n�
k=1
ak − n
�y + n
n�k=1
ayk =
�n�
k=1
ak − n
�z + n
n�k=1
azk =
�n�
k=1
ak − n
�x+ n
.
Mihaly Bencze
286 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25228. If a, b, c ∈ C are given, then solve the following system:(x1 + a)x2 + b = (x2 + a)x3 + b = ... = (xn + a)x1 + b = c.
Mihaly Bencze
PP25229. Let ABC be a triangle and MNK be the triangle for whichMN = ma, NK = mb, KM = mc. Let T ∈ Int (MNK) . Prove that(TM + TN + TK) (TM · TN + TN · TK + TK · TM) ≥≥ TM ·m2
a + TN ·m2b + TK ·m2
c .
Mihaly Bencze
PP25230. Let A1A2...An be a convex polygon, and M ∈ Int (A1A2...An) .
Prove that 12
� MA21+MA2
2−A1A22
MA1·MA2+ n
n−1 ≥ 0.
Mihaly Bencze
PP25231. Denote x1 < 0 < x2 the roots of equationx4 + a1x
3 + b1x2 + c1x+ d1 = 0. Determine all ak, bk, ck, dk ∈ C (k = 1, 2) for
which x1x2 is a root of the equation x6 + x4 + a2x3 + b2x
2 + c2x+ d2 = 0.
Mihaly Bencze
PP25232. Let ABC be a triangle. Solve in R the following system:
�x
(x2+1) cos A2
= y
(y2+1) cos B2
= z(z2+1) cos C
2
xy + yz + zx = 1
Mihaly Bencze
PP25233. Let ABC be a triangle with sides a, b, c. Solve in R the followingsystem:
�x
(x2+1)a= y
(y2+1)b= z
(z2+1)c
xy + yz + zx = 1.
Mihaly Bencze
PP25234. In all triangle ABC holds 8 (�
ma) (�
mamb) ≥ 27�
a2ma.
Mihaly Bencze
Proposed Problems 287
PP25235. If x0, y0, z0 > 0 and 2xn+1 = xn + 1yn; 2yn+1 = yn + 1
zn,
2zn+1 = zn + 1xn
for all n ≥ 1 then the sequences (xn)n≥1 , (yn)n≥1 and(zn)n≥1 are convergent and compute its limits.
Mihaly Bencze
PP25236. If ak, bk ∈ R (k = 1, 2, ..., n) and c ≥n�
k=1
(ak − bk)2 then
vs�n�
k=1
a2k
��n�
k=1
b2k − c
�+
�n�
k=1
b2k
��n�
k=1
a2k − c
�≤ 2
�n�
k=1
akbk
�2
.
Mihaly Bencze
PP25237. If ak ≥ −1 (k = 1, 2, ..., n) , then 9n�
k=1
a3k +3n�
k=1
a2k + n ≥ 5n�
k=1
ak.
Mihaly Bencze
PP25238. Determine all triangles ABC and all points M ∈ (BC) andK ∈ (BC) such that
�AB2 ·MC +AC2 ·MB
� �AB2 ·DK +AC2 ·DB2
�=
=�AB2 ·MC2 +AC2 ·MB2
� �AB2 ·KC +AC2 ·KB
�.
Mihaly Bencze
PP25239. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = 1 then
nn�
k=1ak
≤ �cyclic
11+a1+a2+...+an−1
≤ 1.
Mihaly Bencze
PP25240. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = n then
n�k=1
n−1−a3kak
≥ n (n− 2) .
Mihaly Bencze
288 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25241. If ak > 0 (k = 1, 2, ..., n) then
1).n�
k=1
ak +�
cyclic
a1(a2+...+an)
2 ≥ 2nn−1
2).n�
k=1
a2k +�
cyclic
a21(a2+...+an)
4 ≥ 2n(n−1)2
Mihaly Bencze
PP25242. Prove that12
n�k=1
1k2
+ 12
n�k=1
�12 + 3
4 + ...+ kk+1
�2+
n�k=1
1k
�12 + 1
3 + ...+ 1k+1
�≥ n.
Mihaly Bencze
PP25243. If n, k ∈ N∗ then
12k
+ 13k
+ ...+ 1(n+1)k
≤ n
�1− n
�(2k−1)(3k−1)...((n+1)k−1)
((n+1)!)k
�.
Mihaly Bencze
PP25244. In all triangle ABC holds�
bc�1 + sin A
2
�2 ≤ s2 + 6r2 + 24Rr.
Mihaly Bencze
PP25245. In all triangle ABC holds�
bc sin A2 ≤ 2r (4R+ r) .
Mihaly Bencze
PP25246. In all triangle ABC holds�
(a− b)2 + 3�abc�1 + sin A
2
� �1 + sin B
2
� �1 + sin C
2
�� 23 ≤ 3s2.
Mihaly Bencze
PP25247. If ak ∈ N∗ (k = 1, 2, ..., n) such thatn�
k=1
a2k is a perfect square,
then exist p, q, r ∈ N∗ for which
n�
k=1ak
�
1≤i<j≤n
aiaj
2
= 1p + 2
q +1r .
Mihaly Bencze
Proposed Problems 289
PP25248. If ak ∈ N∗ (k = 1, 2, ..., n) such thatn�
k=1
a2k is a perfect square,
then exist br ∈ N∗ (r = 1, 2, ...,m+ 1) such that
n�
k=1ak
�
1≤i<j≤n
aiaj
m
=m+1�r=1
1br.
Mihaly Bencze
PP25249. If x0 = x1 = 1 and n (n+ 1)�xn+1
√xn−1 − xn
√xn�=
√xn−1xn
for all n ≥ 1, then compute limn→∞
n (4− xn) .
Mihaly Bencze
PP25250. In all triangle ABC holds
1).��
1 + sin A2
�2 ≤ s2
2Rr
2).��
1 + sin A2
�4 ≤ s2(s2−r2−4Rr)8R2r2
3).�
sin A2 ≤ s2+r2+4Rr
4Rr
4).��
1 + sin A2
�≤ s2
4Rr
Mihaly Bencze
PP25251. If a, b, c > 0 then 3�
a2�
ab ≥� 1a+b +
16
��� c(a−b)2
(a+c)(b+c)�
ab
�2
.
Mihaly Bencze
PP25252. In all triangle ABC holds� b2+c2
(2a2+2c2−b2)(2a2+2b2−c2)≤ s2−r2−4Rr
9s2r2.
Mihaly Bencze
PP25253. Determine all ak ∈ N∗ (k = 1, 2, ..., n) for whichn�
k=1
3k2+3k+1a3ka
3k+1
=n(n2+3n+3)
(n+1)3.
Mihaly Bencze
PP25254. Solve in R the following system
2[log2 x1] + x[log2 x3]2 = 2[log2 x2] + x
[log2 x4]3 = ... = 2[log2 xn] + x
[log1 x2]2 = 2 where
[·] denote the integer part.
Mihaly Bencze
290 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25255. Prove thatn−1�k=1
1cosx−cos kπ
n
= cosxsin2 x
− n cos(nx)sinx sinnx .
Mihaly Bencze
PP25256. Let a0, a1, ..., an−1 ∈ N∗ such that a0 + a1 + ...+ an−1 = n. Provethat k0a0 + k1a1 + ...+ kn−1an−1 =
kn−1k−1 if and only if
a0 = a1 = ... = an−1 = 1, when k ≥ 2.
Mihaly Bencze
PP25257. Let 1 = d1 < d2 < ... < dk = n the positive divisors of n. Provethat d1, d2, ..., dk are not in geometrical progression.
Mihaly Bencze
PP25258. In all triangle ABC holds r�s2 − (2R+ r)2
�≤ 4R3.
Mihaly Bencze
PP25259. Determine all functions f : N → N for which
f�x4 − x3f (y) + x2f2 (y)− xf3 (y) + f4 (y)
�=
x5+f(y5)x+y for all x, y ∈ N.
Mihaly Bencze
PP25260. Prove thatn�
k=1
1
(k+2)((2k)!)1k≥ n(n+3)
2(n+1)(n+2) .
Mihaly Bencze
PP25261. Determine all bijective functions f : [0,+∞) → [0,+∞) forwhich f (x+ y + z) =f (x)+ f (y)+ f (z)+2f−1 (f (x) f (y))+2f−1 (f (y) f (z))+2f−1 (f (z) f (x))for all x, y, z ≥ 0.
Mihaly Bencze
PP25262. Solve in Z the equation x3 + y3 = 7z + 2016.
Mihaly Bencze
Proposed Problems 291
PP25263. If ak ∈ N∗\ {1} (k = 1, 2, ..., n) then
logna1 (a2!) + logna2 (a3!) + ...+ lognan (a1!) ≥ nn�
k=1
(ak − 1) .
Mihaly Bencze
PP25264. Determine all xk ∈ N (k = 1, 2, ..., n) for whichn�
k=1
x2k +n�
k=1
xk =n�
k=1
xk.
Mihaly Bencze
PP25265. If xn+m + xn−m = x4n for all n,m ∈ N (n ≥ m) then determinex2016.
Mihaly Bencze
PP25266. In all triangle ABC holds
1).� a+b
b+c ≥ 134 − 3r(r+4R)
4s2
2).� a
b ≥ 4− 3r(4R+s)s2
3).� ra+rb
rb+rc≥ 4− 3s2
(4R+r)2
4).� sin2 A
2+sin2 B
2
sin2 B2+sin2 C
2
≥ 4− 3(s2+r2−8Rr)4(2R−r)2
5).� cos2 A
2+cos2 B
2
cos2 B2+cos2 C
2
≥ 4− 3(s2+(4R+r)2)4(4R+r)2
Mihaly Bencze
PP25267. Prove thatn�
k=1
�2k (k + 1)
√6�> n
√6
3(n+1) where {·} denote the
fractional part.
Mihaly Bencze
PP25268. Solve the equationn�
k=1
�x
k(k+1)(k+2)
�= n, where [·] denote the
integer part.
Mihaly Bencze
292 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25269. In all triangle ABC holds������
a cosA b cosB c cosCa b c1 1 1
������= s
2Rr (a− b) (b− c) (c− a) .
Mihaly Bencze
PP25270. In all triangle ABC holds:1).�
a ma = 2�s2 − r2 − 4Rr
�
2).�
ma wa = s2
3).�
ha wa = 4s2r2��
s2+r2+4Rr4sRr
�2− 1
Rr
�
Mihaly Bencze
PP25271. Solve in R the following system
x21 + 8 = 6x2 +�1 +
√x3 − 2
x22 + 8 = 6x3 +�1 +
√x4 − 2
−−−−−−−−−−−−−x2n + 8 = 6x1 +
�1 +
√x2 − 2
Mihaly Bencze
PP25272. Determine all z ∈ C∗ for which��z + 1
z
�� =��z2n+1 + 1
z2n+1
�� = r ≥ 2
Mihaly Bencze
PP25273. Determine all ak > 0 (k = 1, 2, ..., n) for which holdsn�
k=1
akxakx ≥ n for all x > 0.
Mihaly Bencze
PP25274. Determine all n ∈ N (n ≥ 2) for whichn�
k=1
n
�8 + 1
k > n�
n√5 + n
√7− n
√4�.
Mihaly Bencze
PP25275. Determine all 0 < x1 ≤ x2 ≤ ... ≤ xn for whicharctgx1, arctgx2, ..., arctgxn are in arithmetical progression.
Mihaly Bencze
Proposed Problems 293
PP25276. In all triangle ABC holds� 1
a3
�128R4s2r2 − a4 (b2 + c2 − a2)2 ≥ 3
√2srR .
Mihaly Bencze
PP25277. In all triangle ABC holds� 1
1+sin2 A≤� 1
1+sinA sinB .
Mihaly Bencze
PP25278. In all acute triangle ABC holds� 1
1+cos2 A≤� 1
1+cosA cosB .
Mihaly Bencze
PP25279. Let be 1 = d1 < d2 < ... < dk = n the positive divisors of n.Determine all n for which k = 2016 and n = d21 + d22 + ...+ d22016.
Mihaly Bencze
PP25280. If ak ∈ R (k = 1, 2, ..., n) such that aiaj ≥ 1 (i, j ∈ {1, 2, ..., n})then
n�k=1
1a2k+1
≥ �cyclic
1a1a2+1 . If 0 < aiaj ≤ 1 (i, j ∈ {1, 2, ..., n}) then holds
the reverse inequality.
Mihaly Bencze
PP25281. Let ABCD be a convex quadrilateral, we construct in exteriorthe equilateral triangles ABE,BCF,CDG and DAH. Determine allquadrilaterals ABCD for which the lines AC,BD,EG,FH are concurents.
Mihaly Bencze and Ferenc Olosz
PP25282. If a, b > 1 then solve in R the following systemax1 + b−x2 = ax2 + b−x3 = ... = axn + b−x1 = a+ b
Mihaly Bencze and Gyorgy Szollosy
PP25283. In all triangle ABC holds�� a
ma
�λ≥ 3�
2√3
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
294 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25284. If a > 1 and b ≥ −14 then solve in (0,+∞) the following system
a−bxloga x2
1 + log2a x3 = x4 + loga x5 + b
a−bxloga x3
2 + log2a x4 = x5 + loga x6 + b−−−−−−−−−−−−−−−−−a−bx
loga x1n + log2a x2 = x3 + loga x4 + b
.
Mihaly Bencze and Gyorgy Szollosy
PP25285. Let be P (x) = x4 + ax3 + bx2 + cx+ d when a, b, c, d ∈ R and forwhich P (x) = 1 have integer roots. Exist x0 ∈ Z for which P (x0) = 2016?
Mihaly Bencze
PP25286. In all triangle ABC holds
1).�
ama ≤√3�s2 − r2 − 4Rr
�
2).� ma
bc ≤√3(s2−r2−4Rr)
4sRr
3).�
a2ma ≤√3s�s2 − r2 − 4Rr
�
Mihaly Bencze
PP25287. If f : R → (0,+∞) is a continuous function, then
n�k=1
�
k+1�
k
xf(x)dx
�2
k+1�
k
f2(x)dx
≤ n(n2+3n+3)3 .
Mihaly Bencze
PP25288. If A,B ∈ M2 (R) and λ > 0 thendet�A2 − λB2
�+ λ det (AB −BA) ≤ (detA+ λ detB)2 ≤
≤ det�A2 + λB2
�+ λ det (AB +BA) .
Mihaly Bencze
PP25289. If f, g : [a, b] → R are continuous functions for which:b�xf (t) dt ≥ g (x) for all x ∈ [a, b] and
b�af (x) dx = α and
b�ag (x) dx = β then
1).�b3 − a3
� b�af2 (x) dx ≥ 1
3 (6aα+ 6β − 1)
Proposed Problems 295
2). if a ≥ 0 then�b2√b− a2
√a�2 b�
a|f (x)|3 dx ≥ 1
20 (15aα+ 15β − 2) .
Traian Ianculescu
PP25290. 1). If A,B ∈ M2 (R) then det�A2�−B2 + det (AB −BA) ≤
≤ (detA+ detB)2 ≤ det�A2 +B2
�+ det (AB +BA)
2). If A,B ∈ Mn (R), n ≥ 2 and AB = BA; a, b ∈ R, a2 < 4b thendet�b�A2 +B2
��+�a2 − 2b
�AB − a (A+B) + In ≥ 0
(In conection with problems L.906, L.927, L.1067, RMC)
Traian Ianculescu
PP25291. Solve in R the equation
loga
�a3 loga
�
x+a2
4x
�
12ax2−16x3
�− a
12x2
a2− 16x3
a3 −�x+ a2
4x
�where a > 1. (A generalization
of problem PP. 20281, Octogon Mathematical Magazine).
Traian Ianculescu
PP25292. If a, b, c, x, y > 0 then
1). a+ b+ c+ a(bx+cy)2
+ b(cx+ay)2
+ c(ax+by)2
≥ 6x+y
2). a2 + b2 + c2 +�
a(bx+cy)2
�2+�
b(cx+ay)2
�2+�
c(ax+by)2
�2≥ 3
x2+y2
(A generalization of problem 27070 GMB).
Traian Ianculescu
PP25293. Determine the number of increasing functionsf : {1, 2, ...,m} → {1, 2, ...,m} for which
��f2 (x)− f2 (y)�� ≤��x3 − y3
�� for allx, y ∈ {1, 2, ...,m} , where m ∈ N∗.
Mihaly Bencze
PP25294. If zk ∈ C (k = 1, 2, ..., n) , such thatn�
k=1
|zk| ≤ 1 then����
n�k=1
znk − nn�
k=1
zk
���� ≤ 1.
Mihaly Bencze
296 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25295. Determine all n ∈ N for which exist ak ∈ N∗ (k = 1, 2, ..., n) such
thatn�
k=1
�a1+a2+...+am−1
am
�= n when [·] denote the integer part.
Mihaly Bencze
PP25296. In all triangle ABC holds 4 ≤ 6Rr + 9r(4R+r)
s2.
Mihaly Bencze
PP25297. If 0 < a < b then (2n+ 1) (2m+ 1)�bn+m+1 − an+m+1
�2 ≤≤ (n+m+ 1)2
�b2n+1 − a2n+1
� �b2m+1 − a2m+1
�for all n,m ∈ N.
Mihaly Bencze
PP25298. Solve the equation�
cyclic
(x− a1 − a2 − ...− an−1)n =
n�k=1
ank .
Mihaly Bencze
PP25299. If 0 < a < b thenn�
i=1(nki + 1)
�b1+
n�
i=1ki− a
1+n�
i=1ki
�n
≤�1 +
n�i=1
ki
�n n�i=1
�bnki+1 − anki+1
�,
for all ki ∈ N (i = 1, 2, ..., n).
Mihaly Bencze
PP25300. If en =�1 + 1
n
�nthen compute lim
n→∞nλ�eµn+1 − eµn
�where
λ, µ ∈ R.
Mihaly Bencze
PP25301. Solve in R the following system
�1 + 3
1x1
�(13− 7x2) = 16 (2 + 7x3)
�1 + 3
1x2
�(13− 7x3) = 16 (2 + 7x4)
−−−−−−−−−−−−−−−−�1 + 3
1xn
�(13− 7x1) = 16 (2 + 7x2)
Mihaly Bencze and Gyorgy Szollosy
Proposed Problems 297
PP25302. In all triangle ABC holds�� a
−a+b+c
�λ≥ 3�R2r
�λfor all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP25303. In all triangle ABC holds� −a+b+c
−a+2b+2c ≥ 78 − 2(s2−2r2−8Rr)
5r(4R+r) .
Mihaly Bencze
PP25304. Prove thatn�
k=1
logk+1 (k + 1) (k + 2) ≥�1 + n�ln (n+ 2)
�n.
Mihaly Bencze
PP25305. The triangle ABC is equilateral if and only if�sinA
�2 sin2C − sin2A = s2−r2−4Rr
2R2 .
Mihaly Bencze
PP25306. Solve in Z the equationn�
k=1
(1 + xk) = tn−1n�
k=1
xk.
Mihaly Bencze
PP25307. Determine all functions f : [1,+∞) → R for which
f
�n�
k=1
xk
�=�
cyclic
x1f (x2...xn) for all xk ≥ 1 (k = 1, 2, ..., n) .
Mihaly Bencze
PP25308. Let ABC be a triangle, denote RA the radius of interior circ
tangent in A and tangent to BC etc. Prove that�� 1
RA
�λ≤ 3�
23r
�λfor all
λ ∈ [0, 1] .
Mihaly Bencze
PP25309. In all triangle ABC holds� cos2 B
2+cos2 C
2−cos2 A
2�
(3n−1) cos2 C2+cos2 A
2
≥ 43n+1
�4R+r2R
for all n ∈ N∗.
Mihaly Bencze
298 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25310. In all triangle ABC holds� ctgA
2
2tgA2+tgB
2
≥ 3.
Mihaly Bencze
PP25311. If ak > 0 (k = 1, 2, ..., n), λ ∈ R, b1, b2, ..., bn−1 > 0 thena λ1
aλ2 (b1a1+b2a3+b3a4+...+bn−1an)+
aλ2aλ3 (b1a2+b2a4+b3a5+...+bn−1a1)
+ ...
+ aλnaλ1 (b1an+b2a2+...+bn−1an−2)
≥ n2
(b1+...+bn−1)n�
k=1
ak
. A generalization of problem
3853 Crux Mathematicorum.
Mihaly Bencze
PP25312. In all acute triangle ABC holds
1).� cosA
ha≥ 2(
�
cos A2 )
2
4R+r
2).� cosA
ra≥ 4R(
�
cos A2 )
2
s2+r2+4Rr
Mihaly Bencze
PP25313. Compute limn→∞
�n�
k=1
arctgkλ+k2
− 1λ+1arctg
2n
�when λ > 0.
Mihaly Bencze
PP25314. Let ABC be a triangle. Determine all the numbers x, y ∈ R suchx+ y = 1 for which ax+cy
b + bx+azc + cx+bz
a ≤ Rr + 1.
Mihaly Bencze
PP25315. In all acute triangle ABC holds��� 1
ha+ 1
ra
�cosA
�λ≥ 3
(3r)λ
for all λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP25316. If 1n+1
√xn+1−1 − 1
n√xn−1 = 1 and x1 = 2 then
n�k=1
xk = (n+1)n
n! .
Mihaly Bencze
PP25317. If 0 ≤ a, b, c ≤ 1 and a+ b+ c = 2 then��
a2b� ��
ab2�≤ 1.
Mihaly Bencze
Proposed Problems 299
PP25318. If n, k ≥ 3 then exist p ∈ Z such that(k − 1)n−1 > kp+1−1
k−1 − n ≥ 0
Mihaly Bencze
PP25319. If f, g, h : (−1, 1] → R where f (a) =π�0
(1+cosx) cosxdx1+a2−2a cosx
,
g (a) =π�0
(1+cosx) cosxdx1+a2−2a cos 2x
, h (a) =π�0
(1+cosx) cosxdx1+a2−2a cos 4x
then
1). f (a) + f (−a) ≥ π2). f (a)h (a) = g2 (a) for all a ∈ (−1, 1) .
Gyorgy Szollosy
PP25320. If a, b > 0 then solve the equation ax3 − bx = b3
�b3a .
Gyorgy Szollosy
PP25321. Determine all a ∈ R for which all the roots of the equationx6 − 2 (a+ 1)x5 + 5ax4 − 5a2x2 + 2a2 (a+ 1)x− a3 = 0 are real numbers.
Gyorgy Szollosy
PP25322. Prove that 3−√3 <
3
�√85+92 − 3
�√85−92 <
√3.
Gyorgy Szollosy
PP25323. Solve in R the equation 3x + 35y = 2y − 2
3x = 5.
Gyorgy Szollosy
PP25324. If ε2 + ε+ 1 = 0 and a ∈ C then prove that the roots of theequation z2 − (a− ε) z + a2 + ε = 0 are rational functions of ε.
Gyorgy Szollosy
PP25325. Let be n ∈ N . Prove that the equation
22n �
x8 + y8�2n
= z2 + t2 + w2 have infinitely many integers solutions.
Marius Dragan
300 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25326. Let be In =π�0
n�k=1
cos kxdx. If n ∈ {4k + 1, 4k + 2} then In = 0.
What happens if n ∈ {4k, 4k + 3}?
Gyorgy Szollosy
PP25327. If a, b, c > 0 and a = b+ c then�1 + 1
a2
�a2 �1− 1
a2−b2−c2
�≤�1 + 1
b2
�b2 �1 + 1
c2
�c2.
Marius Dragan
PP25328. Let be x, y, z ∈ N . Find all the rest dividing the numberx3yz + xy3z + xyz3 be 11.
Mihaly Bencze
PP25329. Find the best k ∈ Z such that�√n+
√n+ 1 +
√n+ 2 +
√n+ 3
�=�√
16n+ k�for each n ∈ N, where [·]
denote the integer part.
Marius Dragan
PP25330. If 0 < a ≤ b ≤ c then�a2 + 14
� �b2 + 14
� �c2 + 14
�≥ 26 (3a+ 2b+ c+ 1)2 .
Marius Dragan
PP25331. Solve in N the following equations1). 29x − 20x = y2 2). 29x − 21x = z2
Nicolae Papacu
PP25332. Determine all n ∈ N for which exist xi ∈ N (i = 1, 2, ..., n) such
that x1 < x2 < ... < xn and n+n�
i=1xi =
n�k=1
xi.
Mihaly Bencze
PP25333. Let f : [0, 1] → R a differentiable function for which1�0
f (x) dx = 1 then exist c ∈ (0, 1) for which1�0
f (x) dx = f (c) + cλ−1 (c− 1)
for all λ ∈ R.
Mihaly Bencze
Proposed Problems 301
PP25334. Solve in Z the equation x�y2 + 3z + 2
�= t2.
Mihaly Bencze
PP25335. If xk ∈�0, π2�(k = 1, 2, ..., n) then
2�
cyclic
sin (x1 + x2) ≥ 2n2
n�k=1
sinxk + 2n2
n�k=1
cosxk.
Mihaly Bencze
PP25336. If a > b > c > d > 0 then solve in R the following system:
(c+ d) (ax1 + bx2) = (a+ b) (cx3 + dx4)(c+ d) (ax2 + bx3) = (a+ b) (cx4 + dx5)−−−−−−−−−−−−−−−−−(c+ d) (axn + bx1) = (a+ b) (cx2 + dx3)
Mihaly Bencze
PP25337. If x1 = 2 and xn+1 =�exp�xn−1xn+1
��2for all n ≥ 1 then compute
limn→∞
n (1− xn) .
Mihaly Bencze
PP25338. In all triangle ABC holds� 1
λ+sin2 A≥ 12
4λ+3 for all λ > 0.
Mihaly Bencze
PP25339. If a1 > a2 > ... > a2n > 0 then the function f : R → R when
f (x) =
n�
k=1axk
2n�
p=n+1axp
is injective.
Mihaly Bencze
PP25340. Solve in R the following system
�1 + 2 sin2 x1
� �1 + 2 cos2 x2
�= 4 sin 2x3�
1 + 2 sin2 x2� �
1 + 2 cos2 x3�= 4 sin 2x4
−−−−−−−−−−−−−−−−−�1 + 2 sin2 xn
� �1 + 2 cos2 x1
�= 4 sin 2x2
.
Mihaly Bencze
302 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25341. In all triangle ABC holds� ln(ma)−ln(ha)
ln(mb)−ln(hb)rc ≥ 3s
√r.
Mihaly Bencze
PP25342. Let ABCD be a convex quadrilateral such thatmin {A∡, C∡} ≥ 60◦. Prove that 4BD2 ≥ 3max {AB,AD}max {CB,CD} .
Mihaly Bencze
PP25343. If xk ∈�0, π2�(k = 1, 2, ..., n) , then
n�k=1
1sin2 xk cos2 xk
≤ 3n+
�
1−n�
k=1sin2 xk
�2
n�
k=1sin2 xk
+
�
1−n�
k=1cos2 xk
�
n�
k=1cos2 xk
.
Mihaly Bencze
PP25344. Determine all λ > 0 for which��λ+ 1 + λ
√λ� 1
λ+1+�λ+ 1− λ
√λ� 1
λ+1
�= 2 where [·] denote the integer
part.
Mihaly Bencze
PP25345. Denote ak the greater odd divisor of k. Prove that
1).∞�n=1
12n−1a1a2...an
< e− 1
2).n�
k=1
k!a1a2...ak
≤ 2n − 1
Mihaly Bencze
PP25346. In all triangle ABC holds�
w2a sinA cos2 B
2 ≤ sr(s2+(4R+r)2)8R2 .
Mihaly Bencze
PP25347. Solve in Z the equationx (x+ 2) (x+ 4) (x+ 6) (x+ 8) = y2 − 16.
Mihaly Bencze
PP25348. Determine all integers a, b, c ∈ Z such that� ab+c+1 ≤ 1 ≤� 1
b+c+1 .
Mihaly Bencze
Proposed Problems 303
PP25349. In all triangle ABC holds�
w2a sinA ≤ sr(4R+r)
R .
Mihaly Bencze
PP25350. In all triangle ABC holds�
w2a sinA cos2 A
2 ≤ sr((4R+r)2−s2)4R2 .
Mihaly Bencze
PP25351. In all triangle ABC holds� log2
�
21
sinB +21
sinC
�
sinA ≤ 4Rr .
Mihaly Bencze
PP25352. If xk ≥ 1 (k = 1, 2, ..., n) then�xn log2
�2
n−1
�2x1+x2+...+xn−2
�≤ 2
�1≤i<j≤n
xixj (A generalization of
problem 27061 GMB).
Mihaly Bencze
PP25353. If x0 =3√4 and xn+1 = xn + 1
x2nthen
n�k=1
(xk − 1)3 ≤ 3n(n+3)2 ≤
n�k=1
x3k.
Mihaly Bencze
PP25354. In all triangle ABC holds� (a3+2b3+c3+abc)(a3+c3+abc)
(a3+b3+abc)(b3+c3+abc)≤ 3(s2−3r2−6Rr)
2Rr .
Mihaly Bencze
PP25355. 1). If a > 0, a �= 1, b > 1 and n ∈ N∗ then compute
In (a, b) =b�1b
((1−n)x+ln a)xn−3a1x dx
xn−1a1x+1
2). Determine all continuous functions g :�12 , 2�→ R for which
2�12
�(2−x)ex
x(x2+ex)g (x) dx ≥ ln
4(4√e+1)
e2+4≥
2�12
g2 (x) dx.
Traian Ianculescu
304 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25356. 1). If f : [a, b] → R is continuous and a > 0, α ∈ R, α �= 1, then
exist c ∈ (a, b) such that (α− 1) (ab)α−1b�af (x) dx =
�bα−1 − aα−1
�cαf (c)
2). If g : [a, b] → R is a continuous function and G (x) =x�0
g (t) dt, x ∈ [a, b]
and β < 0 then exists c1, c2 ∈ (0, b) , c1 �= c2 such that
1−βb1−β
b�0
G (x) dx = cβ+11 g (c2)
Traian Ianculescu
PP25357. If 0 < a < 1 < c < d then determine all differentiable functionwith continuous derivate for whichf (a · cx + b)− f (a · dx + b) = a (f (cx)− f (dx)) for all x ∈ R.
Mihaly Bencze
PP25358. Solve in Z the equation 2x
y+1 + 3y
2x+1 = 2.
Mihaly Bencze
PP25359. If a0 =13√4
and an+1
�1 + a2n
�= an for all n ≥ 1 the
a242 ∈�
110 ,
19
�and lim
n→∞3√nan = 1
3√3
Traian Ianculescu
PP25360. Prove thatn�
k=1
�2n+2−k
k ≥ n�1 +
n√n!
n+1
�.
Traian Ianculescu
PP25361. If A ∈ M2 (R) and x ∈ R such that det�A2 − x2I2
�≥ 0 then
1). If detA+ x2 ≥ 0 thendet (An − xnI2) + det (An + xnI2) ≥ 22−n |xTr (A)|n for all n ∈ N
2). If A ∈ M2 (Q) and det�A2 − 2I2
�= 0 then A2 = 2I2.
Traian Ianculescu
PP25362. If ak, xk > 0 (k = 1, 2, 3, 4) then� x1a2a3
(x1+x2)(x2+x3)≤ (a1+a2+a3)
2
4(x1+x2+x3).
Traian Ianculescu
Proposed Problems 305
PP25363. If ai > 0 (i = 0, 1, ..., n) are in arithmetical progression with ratio
r > 0 thenn�
i=1
1ai
< nr
�1− n
�a0an
�. (A generalization of problem C.1107
GMB).
Traian Ianculescu
PP25364. If ak > 0 (k = 1, 2, ..., n) then�
cyclic
an2an1+an2+...+ann−1+a1a2...an
≥ 1 (A
generalization of problem C.3201 GMB).
Traian Ianculescu
PP25365. In all triangle ABC holds
1).�
aα�b2α + c2α − a2α
�≤ 3 (abc)α
2).� √
ara
≤ 3�
Rs when α ∈ [0, 1]
Traian Ianculescu
PP25366. If xi > 0 (i = 1, 2, ..., n) and f (λ) =�
(xλ1+xλ
2+...+xλk)
n�
i=1xλi
then
f (λ+ 1) ≥ f (λ) for all λ > 0, where k ∈ {1, 2, ..., n} .
Mihaly Bencze
PP25367. Solve in Z the equation 2x+1 + 3
y2+2+ 5
z3+4= 3.
Mihaly Bencze
PP25368. If x, y, z > 0 and x+ y + z = 1 then� x4
x2+y+z≥ 1
21 .
Mihaly Bencze
PP25369. If ak > 0 (k = 1, 2, ..., n), then�
cyclic
an1a2
≥� a1a2...an−1.
Mihaly Bencze
PP25370. If a, b, c > 0 and λ ∈ (0, 5) then� a3
b(a+λc) ≥5−λ9
�a.
Mihaly Bencze
306 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25371. Let be aij > 0 (i = 1, 2, ..., n; j = 1, 2, ...,m) and xj > 0(j = 1, 2, ...,m) and f : R → (0,+∞) is a convex function. Prove thatm�j=1
fn+1(xij)a1ja2j ...anj
≥ mn+1
n�
i=1
�
m�
j=1aij
�fn+1
�1m
m�j=1
xj
�.
Mihaly Bencze
PP25372. If xk > 0 (k = 1, 2, ..., n) andn�
k=1
xk = 1 then
�cyclic
x31
x21+x2+...+xn
≥ 1n2−n+1
.
Mihaly Bencze
PP25373. In all triangle ABC holds� cos4 A
2
cos2 B2
≥ 4R+r2R
�2((4R+r)2−s2)(4R+r)2+s2
.
Mihaly Bencze
PP25374. If ak > 0 (k = 1, 2, ..., n) , then� a1a3+a22
(a1+a3)a2≥ n.
Mihaly Bencze
PP25375. Let f : R → (0,+∞) be a convex function and xk, ak > 0(k = 1, 2, ..., n). Determine all λ > 0 for whichn�
k=1
fλ(xk)ak
≥ n2
n�
k=1
ak
fλ
�1n
n�k=1
xk
�.
Mihaly Bencze
PP25376. If ak > 0 (k = 1, 2, ..., n) thenn�
k=1
ak +�
cyclic
(a2−a3)2
a1≥�
cyclic
|a1 − a2| .
Mihaly Bencze
PP25377. In all triangle ABC holds�
tg2A2 ctgB2 ≥ 4R+r
s2
�(4R+ r)2 − 2s2.
Mihaly Bencze
Proposed Problems 307
PP25378. If ak > 0 (k = 1, 2, ..., n) then�
cyclic
a21a2
≥�
n�k=1
ak
�����
n�
k=1a2k
�
cyclica1a2
.
Mihaly Bencze
PP25379. If a, b, c > 0 , a �= b and In =π�0
cosnxdxa2+b2−2ab cosx
for all n ∈ N, then�n≥0
In is convergent and compute its limit.
Gyorgy Szollosy
PP25380. If ak > 0 (k = 1, 2, ..., n) and f (u, v) =�
cyclic
au1av2
then
f (u+ 1, v + 1) ≥ f (u, v) for all u, v > 0.
Mihaly Bencze
PP25381. In all triangle ABC holds��−a+b+c
a > 2√2.
Mihaly Bencze
PP25382. Determine the rest of divisor of number 52016 with 2016.
Mihaly Bencze
PP25383. If a > 0, x0 > −1 and 3xn+1 (xn + 1)2 = a+ 2x3n + 3x2n − 1 thencompute lim
n→∞n (1− xn) .
Mihaly Bencze and Gyorgy Szollosy
PP25384. If xn = |zn + z−n| , when z ∈ C∗ and if x1 > 2 thenxn > xn−1 + x2 − x1 for all n ≥ 2.
Mihaly Bencze
PP25385. If A,B ∈ Mn (C) such that ABk = A−B for all n ∈ N∗.Determine all k ∈ N for which In +B is invertible. Determine all k ∈ N forwhich AB = BA.
Mihaly Bencze
308 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25386. Solve in Z the following equation1
xy+z−1 + 1yz+x−1 + 1
zx+y−1 = 3n2−n+1
where n ∈ N∗ is given.
Mihaly Bencze
PP25387. Prove thatn�
k=1
n(n−1)...(n−k+1)knk+1 = 1.
Mihaly Bencze
PP25388. In all triangle ABC holds�
tg2A2 tg2B2 +�
ctg2A2 ctg2B2 ≥ 10.
Mihaly Bencze
PP25389. Prove that for all n, k ∈ N∗ exists m1,m2, ...,mk ∈ N∗ numbers
such that 1 + pk−1n(p−1) =
k�i=1
�1 + 1
mi
�for all p ∈ N, p ≥ 2.
Mihaly Bencze
PP25390. Solve in R the following system:
√x1 − 2 +
�x22 − 4x3 + 3 =
�(x4 − 2)3
√x2 − 2 +
�x23 − 4x4 + 3 =
�(x5 − 2)3
−−−−−−−−−−−−−−−−−−√xn − 2 +
�x21 − 4x2 + 3 =
�(x3 − 2)3
Mihaly Bencze
PP25391. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3.
Prove that a�
bc2(a+b)+5c + b
�ca
2(b+c)+5a + c�
ab2(c+a)+5b <
�3332 .
Jose Luis Dıaz-Barrero
PP25392. Let a, b, c be three positive real numbers such thata2 + b2 + c2 = 9. Prove that b−1
(5√a+2
√b)2
+ c−1
(5√b+2
√c)2
+ a−1
(5√c+2
√a)2
≥ 149
Jose Luis Dıaz-Barrero
PP25393. Let n ≥ 0 be an integer number. Calculate limx→+∞
xlnx
∞�n=0
12016n+x .
Jose Luis Dıaz-Barrero
Proposed Problems 309
PP25394. Let n ≥ 0 be an integer number. Let n be a positive integer.
Show that 1n
n�k=1
�k2k
�nk
��2 ≥ 19
�32
�2n
Jose Luis Dıaz-Barrero
PP25395. If a ∈�0, 14�then solve in R the following system:
x21 + 2ax2 +116 + a =
�a2 + x3 − 1
16
x22 + 2ax3 +116 + a =
�a2 + x4 − 1
16
−−−−−−−−−−−−−−−−x2n + 2ax1 +
116 + a =
�a2 + x2 − 1
16
.
Mihaly Bencze
PP25396. Solve in R the following system:a = x21 +
√a+ x2 = x22 +
√a+ x3 = ... = x2n +
√a+ x1 when a ∈ R is given.
Mihaly Bencze
PP25397. Solve in R the following system:�x1 + 3− 4
√x2 − 1 +
�x2 + 8− 6
√x3 − 1 =
�x2 + 3− 4
√x3 − 1 +�
x3 + 8− 6√x4 − 1 = ... =
�xn + 3− 4
√x1 − 1+
�x1 + 8− 6
√x2 − 1 = 1.
Mihaly Bencze
PP25398. Solve in R the following system:
|x1 + 1|+ 3 |x2 − 1| = x3 + 2 + |x4|+ 2 |x5 − 2||x2 + 1|+ 3 |x3 − 1| = x4 + 2 + |x5|+ 2 |x6 − 2|−−−−−−−−−−−−−−−−−−−−−|xn + 1|+ 3 |x1 − 1| = x2 + 2 + |x3|+ 2 |x4 − 2|
.
Mihaly Bencze
PP25399. Determine all x ∈ R for whichn�
k=1
[kx]3 =�[nx]([nx]+[x])
2
�2for all
n ∈ N∗, when [·] denote the integer part.
Mihaly Bencze
310 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25400. If ak ∈ R∗ (k = 1, 2, ..., n) such thatn�
k=1
ak = 1 andn�
k=1
{ak} = 1.
Prove thatn�
k=1
{amk } = 1 for all m ∈ N, where {·} denote the fractional part.
Mihaly Bencze
PP25401. Determine all λ ∈ R for which the equation sinx = λx haveexactly 2016 solutions.
Mihaly Bencze
PP25402. Prove that x is integer number if and only ifn�
k=1
[kx]2 = n([nx]+[x])([2nx]+[x])6 for all n ∈ N∗, where [·] denote the integer
part.
Mihaly Bencze
PP25403. In all triangle ABC holds
� a(a+b)(a+c)2a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b) ≤
34 .
Mihaly Bencze
PP25404. Solve in R the following system:
log3 (1 + 2x1) = log2 (1 + x2)log3 (1 + 2x2) = log2 (1 + x3)−−−−−−−−−−−−log3 (1 + 2xn) = log2 (1 + x1)
.
Mihaly Bencze
PP25405. If ap > 1 (p = 1, 2, ..., n) and k ∈ {1, 2, ..., n} thenlog2a1+a2+...+ak
(a2+a3+...+ak+1)
a3+a4+...+ak+2+
log2a2+a3+...+ak+1(a3+a4+...+ak+2)
a4+a5+...+ak+3+ ...
+log2an+a1+...+ak−1
(a1+a2+...+ak)
a2+a3+...+ak+1≥ n2
kn�
p=1ap
.
Mihaly Bencze
Proposed Problems 311
PP25406. If (xn)n≥1 is a real sequence such that |xn+1 − xn| ≤ 1 andnyn = x1 + x2 + ...+ xn for all n ∈ N∗, then |yn+1 − yn|+ |yn+2 − yn+1| ≤ 1for all n ∈ N∗.
Mihaly Bencze
PP25407. If An =
������
cos2 nx cos2 ny cos2 nzsin 2nx sin 2ny sin 2nzcos 2nx cos 2ny cos 2nz
������where x, y, z ∈ R then
n�k=1
|Ak| < 2n.
Mihaly Bencze
PP25408. Solve in N the equation�2n
n
�= 2k, when [·] denote the integer
part.
Mihaly Bencze
PP25409. Solve in N the equation�3n
n
�= 3k, when [·] denote the integer
part.
Mihaly Bencze
PP25410. Solve in N the equation�2n+3m
nm
�= 6k, when [·] denote the
integer part.
Mihaly Bencze
PP25411. Let a1, a2, ..., an2 be an arithmetical progression, and let be the
following tabel:
��������
a1 a2 ... anan+1 an+2 ... a2n... ... ... ...a(n−1)2+1 a(n−1)2+2 ... an2
��������. Prove that if we elimine
n elements from this tabel in the rest n (n− 1) elements exist n numbers inarithmetical progression.
Mihaly Bencze
312 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25412. If xk ∈ R (k = 1, 2, ..., n) such thatn�
k=1
sinxk = 0, then
n�k=1
|cosx− sinxk| ≤ n for all x ∈ R.
Mihaly Bencze
PP25413. Denote An the set of every integers of form ±1± 2± ...± n. Byexample A2 = {−3,−1, 1, 3} and A3 = {−6,−4,−2, 0, 2, 4, 6} . Denote |An|the number of elements of the set An. Compute
∞�n=1
1|An| .
Mihaly Bencze
PP25414. If a, b ∈ C then (|1 + ab|+ |a+ b|)2 ≥���(1 + ab)2 − (a+ b)2
��� .
Mihaly Bencze
PP25415. If A ∈ M2 (R) then83 det
�A2 +A+ I2
�≥ (1− detA)2 + (1 + TrA)2 .
Mihaly Bencze
PP25416. If A ∈ Mn (R) is an antisimmetric matrice(aij + aji = 0 for all i, j ∈ {1, 2, ..., n}) then for all xk ≥ 0 (k = 1, 2, ...,m)
holdsm�k=1
det (A+ xkIn) ≥�det
�A+ m
�m�k=1
xkIn
��m
.
Mihaly Bencze
PP25417. In all triangle ABC holds�� ab
(b+c)(c+a) ≤�
6(s2−r2−Rr)s2+r2+2Rr
.
Mihaly Bencze
PP25418. In all triangle ABC holds s2+(4R+r)2
4sR − 3s2(4R+r) ≥
√3.
Mihaly Bencze
PP25419. Solve in Z the equation 1x + 1
y+2 + 1z+3 = 3.
Mihaly Bencze
Proposed Problems 313
PP25420. If x1 = 2 and x2n+1 = xn + 1n for all n ≥ 1 then compute
limn→∞
n (1− xn) .
Mihaly Bencze
PP25421. In all triangle ABC holds� a(a+b)3(a+c)
(b+c)(b(a+b)3+(a+c)c3+7(a+c)(a+b)3)≤ 1
3 .
Mihaly Bencze
PP25422. Prove that
n�
k=1(−1)k−1 1
k2(nk)
Hn
Hn
≥n�
k=1
(Hk)1k when Hn =
n�k=1
1k .
Mihaly Bencze
PP25423. Let G = (0,+∞) and (log3 (x ◦ y))n = (log3 x)n + (log3 y)
n − 3n
where n = 2k + 1, k ∈ N∗ and for all x, y ∈ G
1). Prove that (G, ◦) is Abelian group
2). Prove that (G, ◦) ∼= (R,+)
Mihaly Bencze
PP25424. If Hn = 1 + 12 + ...+ 1
n then
n�k=1
�1 + 1
k
�nk
k+1�0
xn−1 ln�
kk+1 − x
�dx = − (Hn + ln (n+ 1)) .
Mihaly Bencze
PP25425. If ak > 0 (k = 1, 2, ..., n), then
�cyclic
a1
�1aλ2
+ 1aλ3
+ ...+ 1aλn
�≥
n�
k=1ak
(n−1)λ−1
n�
k=1
1ak
− nn�
k=1
ak
λ
for all
λ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP25426. If a > 1, b ≥ −14 then solve in (0,+∞) the equation
a−bxloga x + log2a x = x+ logb x+ b.
Gyorgy Szollosy
314 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25427. If Hn = 1 + 12 + ...+ 1
n thenn�
k=0
Hkk ≥ 1
2 (Hn)2 + ln2 n
2n .
Mihaly Bencze
PP25428. Solve in R the equation 9x2 − 9x− 2 = 12 sinπx.
Gyorgy Szollosy
PP25429. Solve in R the equation�65
� 1x + 2
1+( 95)
x−1 = 3.
Gyorgy Szollosy
PP25430. If n,m ∈ N (n ≥ 2) , α, ak > 0 (k = 1, 2, ..., n) such thatn�
k=1
aαk ≤ n then 1n + m
n�
k=1ak
≥ 1+mnn�
k=1ak
.
Gyorgy Szollosy
PP25431. Solve in R the equation 3x + 35y = 2y − 2
3x = 5.
Gyorgy Szollosy
PP25432. In all scalene triangle ABC holds�� b−a
ma−mb− 2
3
�2+
32(s2+r2−2Rr)s2+r2+2Rr
< 64�
s2−r2−Rrs2+r2+2Rr
�2.
Mihaly Bencze
PP25433. In all scalene triangle ABC holds�� b−a
ma−mb− 2
3
��c−b
mb−mc− 2
3
�≤ 16(s2+r2−2Rr)
s2+r2+2Rr.
Mihaly Bencze
PP25434. In all scalene triangle ABC holds�b−a
ma−mb− 2
3
��c−b
mb−mc− 2
3
��a−c
mc−ma− 2
3
�≤ 128Rr
s2+r2+2Rr.
Mihaly Bencze
PP25435. In all scalene triangle ABC holds�� b−a
ma−mb
�2≥ 16
27
�� ma+mba+b
�2.
Mihaly Bencze
Proposed Problems 315
PP25436. In all scalene triangle ABC holds� b−a
ma−mb<
2(7s2−r2+2Rr)3(s2+r2+2Rr)
.
Mihaly Bencze
PP25437. Denote F the area of the triangle with sides cos A2 , cos
B2 , cos
C2
when ABC is a given triangle. Prove that
16F 2 ≤ cos A2 cos B
2 cos C2
�cos A
2 + cos B2 + cos C
2
�.
Mihaly Bencze
PP25438. In all triangle ABC holds3�s2 − r2 − 4Rr
�2 − 12s2r2 ≥�
mamb (a− b)2 .
Mihaly Bencze
PP25439. In all triangle ABC holds� (a+b)2m2
b+(a+c)m2c
2a+b+c ≥�mamb.
Mihaly Bencze
PP25440. In all triangle ABC holds7(s2−r2−4Rr)
4 ≥� ama ≥ s(s2+r2−2Rr)2R .
Mihaly Bencze
PP25441. In all triangle ABC holds 3�
(2a+ b+ c) (mbmc)2 +
916
�(b+ c) (b+ c− a) (b− c)2
�a2 + b2 + c2 + 3a (b+ c)
�≥
≥��
(a+ b)m2b +�
(a+ c)m2c
�2.
Mihaly Bencze
PP25442. In all triangle ABC holds
6�
(b− c) (bmc − cmb) (bmc + cmb) ≥���
(b+ c) (a2 + b2 + c2) |b− c|�2
.
Mihaly Bencze
PP25443. If Sn =n�
k=1
2k−1�
k2
k+1
�when [·] denote the integer part, then
�2−n−1Sn
�= n
2 − 1 for all n = 2k.
Mihaly Bencze
316 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25444. Let A1A2...An be a convex polygon. Prove that
n+�
cyclic
�a1
a2+...+an
�1 +�
a2+...+ana2+...+an−a1
���=�
cyclic
�a2+...+an
a2+...+an−a1
�when [·]
denote the integer part.
Mihaly Bencze
PP25445. Let ABCD be a convex quadrilateral such that A∡ ≥ 90◦,C∡ ≥ 90◦. Denote E respective F the proiection of A and C to BD. Provethat 1
AE·CF ≥ 1AB·DC + 1
AD·BC .
Mihaly Bencze
PP25446. Prove thatm�
n=1
1n�
k=1[√k2−k+1+
√k2+k+1]
= mm+1 .
Mihaly Bencze
PP25447. Let a, b, c > 0 then determine all λ ∈ R for which� ab ≥ λ
� abbλ+ca
.
Mihaly Bencze
PP25448. In all triangle ABC holds� a+b
c
��b+c−a
a +�
c+a−bb
�≤�s2+r2−2Rr
2Rr
�2− 2(s2−r2−Rr)
Rr .
Mihaly Bencze
PP25449. If a1 = 1 and ak > 0 (k = 1, 2, ..., n) such that
1a1+a2
+ 1a2+a3
+ ...+ 1an−1+an
= an − 1 for all n ≥ 1 thenn�
k=1
1a2ka
2k+1
= nn+1 .
Mihaly Bencze
PP25450. In all triangle ABC holds
1).��
a2 + b2
3 + c2�≥ 8s2Rr
�s2 + r2 + 2Rr
�
2).��
r2a +r2b3 + r2c
�≥ 4s4Rr2
3).���
sin A2
�4+ 1
3
�sin B
2
�4+�sin C
2
�4� ≥ r2((2R−r)(s2+r2−8Rr)−2Rr2)512R5
Mihaly Bencze
Proposed Problems 317
PP25451. Let A1A2...An be a convex polygon. Prove that�
cyclic
√an(a1+a2+...+an−1−an)
a1+a2+...+an−1≤ n
2 .
Mihaly Bencze
PP25452. In all triangle ABC holds� r4a
λr2br2c+((4R+r)2−2s2)
2 ≥ 3λ+3 for all
λ > 0.
Mihaly Bencze
PP25453. If ak > 0 (k = 1, 2, ..., n) and λ > 0 then
�cyclic
1
λa1+2n�
k=1ak
≤ 1(n+1)2
�1λ
n�k=1
+n�
cyclic
1a1+a2
�.
Mihaly Bencze
PP25454. In all triangle ABC holds� rarb
5ra+2(ra+rc)≤ 4R+r
12 + s2+r(4R+r)64R .
Mihaly Bencze
PP25455. In all triangle ABC holds� (s−a)(s−b)
5(a+c)−b ≤ s24 +
r(s2+(4R+r)2)8sR .
Mihaly Bencze
PP25456. If ak, bk ∈ R (k = 1, 2, ..., n) , then�
n�k=1
a2k
��n�
k=1
b2k
�≥�
n�k=1
akbk
�2
+ 2n(n−1)
��
1≤i<j≤n|aibj − ajbi|
�2
.
Mihaly Bencze
PP25457. If ak ∈ [0, 1] (k = 1, 2, ..., n) then�
cyclic
a1a31+2(a2+...+an)+n+1
≤ 13 .
Mihaly Bencze
PP25458. If x, ak > 1 (k = 1, 2, ..., n) , then loga1a2...an x ≤ n2n�
k=1
logak x.
Mihaly Bencze
318 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25459. Determine an (n ≥ 1) if a1 = 1 and1
a1a3a5+ 1
a3a5a7+ ...+ 1
a2n−1a2n+1a2n+3= n(n+2)
3a2n+1a2n+3.
Mihaly Bencze
PP25460. Determine an (n ≥ 1) if a1 = 1 and
14
a1a3+ 24
a3a5+ ...+ n4
a2n−1a2n+1=
n(n+1)(n2+n+1)6a2n+1
.
Mihaly Bencze
PP25461. Determine all function f : N → N for which(f(n3)+1)(f3(n)−1)
(f(n2)−n+1)(f2(n)+n+1)= f (n− 1) (f (n) + 1) for all n ∈ N∗.
Mihaly Bencze
PP25462. In all triangle ABC holds1).� a2
a4+b2c2≤ 1
4R
2).� r2a
r4a+r2br2c≤ 4R+r
2s2r
3).� (sin A
2 )4
(sin A2 )
8+(sin B
2sin C
2 )4 ≤ 4R(2R−r)
r2
4).� (cos A
2 )4
(cos A2 )
8+(cos B
2cos C
2 )4 ≤ 4R(4R+r)
s2
Mihaly Bencze
PP25463. If ak > 0 (k = 1, 2, ..., n) , then
�cyclic
(n−1)an−11 +a2a3...an
an−22
≥ n2 n
�n�
k=1
ak.
Mihaly Bencze
PP25464. Solve in R the following system:
[x1] +�x2 +
13
�+�x3 +
23
�= [3x4]
[x2] +�x3 +
13
�+�x4 +
23
�= [3x5]
−−−−−−−−−−−−−−−[xn] +
�x1 +
13
�+�x2 +
23
�= [3x3]
, when [·] denote the integer part.
Mihaly Bencze
Proposed Problems 319
PP25465. Solve in R the equation�x2�+�x+ 1
2
�= [2
√x] , when [·] denote
the integer part.
Mihaly Bencze
PP25466. If a1 = 0 and (an+1 − an) (an+1 − an − 2) = 4an for all n ≥ 1
thenn�
k=2
1ak
= n−1n .
Mihaly Bencze
PP25467. Solve in R the following system
�x21 − x2 − 2
�= [x3]�
x22 − x3 − 2�= [x4]
−−−−−−−−−�x2n − x1 − 2
�= [x2]
, when
[·] denote the integer part.
Mihaly Bencze
PP25468. Solve in R the following equation
n {x}+n�
k=1
�x+ 1
k
�= 1 +
n�k=1
{kx} , when {·} denote the fractional part.
Mihaly Bencze
PP25469. If Sn =n�
k=1
�k3
�when [·] denote the integer part, then compute
∞�n=3
1S2n.
Mihaly Bencze
PP25470. Determine all a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for which��ab�2n −
�cd�2n���
cd�2n
+�cd�2n+1
+ 1�is divisible by abcd.
Mihaly Bencze
PP25471. If a1 =16 and (n+ 3) an+1 = (n+ 1)
�an + 1
2
�for all n ≥ 1, then
computen�
k=1
[ak] when [·] denote the integer part.
Mihaly Bencze
320 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25472. In all triangle ABC holds� 4
�
tgA2tgB
2
1+ 4�
9tgB2tgC
2
≥ 16 .
Mihaly Bencze
PP25473. Determine all ak ∈ R (k = 1, 2, ..., n) for which a1 = 1 andn�
k=1
a3k =
�n�
k=1
ak
��n�
k=1
a2kk
�
Mihaly Bencze
PP25474. If xk ∈ R (k = 1, 2, ..., n) ,λk > 0 (k = 1, 2, ..., n) such thatn�
k=1
λk = 1, then |λ1x1 + λ2x2 + ...+ λnxn|+ |λ1x2 + λ2x3 + ...+ λnx1|+
...+ |λ1xn + λ2x1 + ...+ λnxn−1| ≤n�
k=1
|xk| .
Mihaly Bencze
PP25475. Solve in R the following system:
�1
10−x1
�= 1
10−[x2]�1
10−x2
�= 1
10−[x3]
−−−−−−−−�1
10−xn
�= 1
10−[x1]
when [·]
denote the integer part.
Mihaly Bencze
PP25476. Solve in Z the equation 1x2+1
+ 2y3+3
+ 3z4+5
= 3.
Mihaly Bencze
PP25477. In all triangle ABC hold� (r2a+s2)(r2b+s2)
r2a+r2b+2s2≥ 2.
Mihaly Bencze
PP25478. Determine all function f : N∗ → N∗ for whichn�
k=1
k4
f(k) =n4f2(n+1)4f2(n)
for all n ∈ N∗.
Mihaly Bencze
Proposed Problems 321
PP25479. Solve in R the following system:
�4x1+44x2−5
�= 2x3−1
3�4x2+44x3−5
�= 2x4−1
3
−−−−−−−−�4xn+44x1−5
�= 2x2−1
3
, when [·]
denote the integer part.
Mihaly Bencze
PP25480. Solve in R the following system:�x21�− 4 [x2] =
�x22�− 4 [x3] = ... =
�x2n�− 4 [x1] = −3, when [·] denote the
integer part.
Mihaly Bencze
PP25481. Compute S (p, r, n) =n�
k=1
pk�
kr
k+1
�, when [·] denote the integer
part. We have S (2, 2, n) = n · 2n − 2n+1 + 2.
Mihaly Bencze
PP25482. Solve in R the following system
1[x1]
+ 1{x2} = 2016x3
1[x2]
+ 1{x3} = 2016x4
−−−−−−−−−−1
[xn]+ 1
{x1} = 2016x2
, when
[·] and {·} denote the integer respective fractional part.
Mihaly Bencze
PP25483. Determine (an)n≥1 such that a1 = 1 andn�
k=1
ka2k = 14n
2a2n+1 for
all n ≥ 1.
Mihaly Bencze
PP25484. If xk > 0 (k = 1, 2, ..., n) and m ∈ N, then�
cyclic
xm+21 +xm+2
2
xm+11 +xm+1
2
+�
cyclic
xm1 +xm
2
xm−11 +xm−1
2
≤ 2�
cyclic
xm+11 +xm+1
2xm1 +xm
2.
Mihaly Bencze
322 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25485. In all triangle ABC holds� tg2Atg3A+tgA
+� ctg2A
ctg3B+ctgA≥� sin2 2A
1+cos2 2A.
Mihaly Bencze
PP25486. In all triangle ABC holds�(tgA+ ctgA)
�tg2A
tg3A+ctgA+ ctg2A
ctg3A+tgA
�≤ 6.
Mihaly Bencze
PP25487. In all triangle ABC holds 27s2(9t2−1)
≤� 1(st)2−(rarb)
2 ≤ 3t2−2(st)2(t2−1)
for all t > 1.
Mihaly Bencze
PP25488. If x ∈�0, π2�then
�sin2 x
�1 +�
1cos2 x
���+�cos2 x
�1 +�
1sin2 x
���+ 2 =
�1
sin2 x
�+�
1cos2 x
�, when
[·] denote the integer part.
Mihaly Bencze
PP25489. In all triangle ABC holds
3 +�
cyclic
�a
b+c
�1 +�
b+cb+c−a
���=�
cyclic
�b+c
b+c−a
�, when [·] denote the integer
part.
Mihaly Bencze
PP25490. If a1 = 3 then determine (an)n≥1 such thatn�
k=1
(2k−1)(2k+1)a22k−1a
22k+1
= n−13a2n+1
for all n ≥ 2.
Mihaly Bencze
PP25491. If (an)n≥1 is an arithmetical progression with ratio r and a1 ≥ 12 ,
then compute the integer part of the expression�
cyclic
3
�1 + r2
a1a2a3.
Mihaly Bencze
Proposed Problems 323
PP25492. Determine all n, k, r, s ∈ N and all prime p and q such thatn4 + (4k + 1)n2 + 4k2 + 4k + 3 = pr + qs.
Mihaly Bencze
PP25493. Solve in R the following system
x1
�n�
k=1
xk
�= 13
x2
�n�
k=1
xk
�= 23
−−−−−−−−xn
�n�
k=1
xk
�= n3
and prove
thatn�
k=1
3
�x2k =
3
�n2(n+1)2(2n+1)3
108 .
Mihaly Bencze
PP25494. In all triangle ABC holds� tgA
2tgB
2
1+λtgC2
�
tgA2tgB
2
≥ 3λ+3 for all λ > 0.
Mihaly Bencze
PP25495. In all triangle ABC holds���
rarbs2
�2+�rbrcs2
�2+�
s2
rcra
�2≥ 3
√3.
Mihaly Bencze
PP25496. Denote f (n) the number of triples (a, b, c) with properties1 ≤ a ≤ b ≤ c ≤ n when a, b, c are in geometrical progression with ratio a
natural number. Prove that∞�k=1
1f2(k)
> π2
24 .
Mihaly Bencze
PP25497. Solve in N the equation 2015x(x+1) +
2016y(y+1) +
2017z(z+1) = 1.
Mihaly Bencze
PP25498. Compute limx→∞
x
�1
ln 2016 − xlnx
∞�n=0
12016n+x
�
Mihaly Bencze and Jose Luis Dıaz-Barrero
324 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25499. If a, b, c > 0 and a+ b+ c = 1, then 111 +
� a2
5a+2 ≥� (a+b)2
5(a+b)+4
Mihaly Bencze and Jose Luis Dıaz-Barrero
PP25500. If ak > 0 (k = 1, 2, ..., n) andn�
k=1
ak = 1 thenn�
k=1
a2k5ak+2 ≥ 1
2n+5 .
Mihaly Bencze and Jose Luis Dıaz-Barrero
PP25501. In all triangle ABC holds� ab
5b+2(a+b) ≤s6 +
(s2+r2+4Rr)2+8s2Rr
32s(s2+r2+2Rr).
Mihaly Bencze
PP25502. In all triangle ABC holds� 1
ra+ 1
ha
3
�
1+ r2
�
1rb
+ 1hb
�
− r2
�
1rd
+ 1hd
�
≥ 2r .
Mihaly Bencze
PP25503. In all triangle ABC holds 2s2
3s2−r2−4Rr+� w2
abc ≤ 3.
Mihaly Bencze
PP25504. In all triangle ABC holds� (ha)
32
(ha)32√ha+2hb
+� (rb)
32
(ra)32√ra+2rb
≥ 2√r.
Mihaly Bencze
PP25505. In all tetrahedron ABCD holds1).� 3√hbhd
ha3√
hbhd+r(hd−hb)≥ 1
r
2).� 3
√rbrd
ra3√
rbrd+r(rd−rb)≥ 1
r
Mihaly Bencze
PP25506. In all triangle ABC holds� r6a
((4R+r)2+9r2a)2 ≥ (4R+r)2
972 .
Mihaly Bencze
PP25507. In all triangle ABC holds� (s−a)6
(a2+9(s−a)2)2 ≥ s2
972 .
Mihaly Bencze
Proposed Problems 325
PP25508. In all triangle ABC holds
1).� cos A
2cos B
2
cos A2+cos B
2−cos C
2
≥� cos A2
2).� mamb
ma+mb−mc≥�ma
Mihaly Bencze
PP25509. In all triangle ABC holds� c2−(a−b)2
3c−a−b ≥ 2s.
Mihaly Bencze
PP25510. In all triangle ABC holds� (cos A
2+cos B
2 )2
(cos A2+cos B
2 )2−cos2 C
2
≥ 4.
Mihaly Bencze
PP25511. In all triangle ABC holds� a2+b2+3c2+8mamb
4c2−a2−b2+8mamb≥ 4.
Mihaly Bencze
PP25512. In all triangle ABC holds� (a+b)λ
(a+b)λ−cλ≥ 3·2λ
2λ−1for all λ ≥ 1.
Mihaly Bencze
PP25513. Let A1A2...An be a convex polygon. Prove that� (a1+a2+...+an−1)λ
(a1+a2+...+an−1)λ−aλn
≥ n(n−1)λ
(n−1)λ−1for all λ ≥ 1.
Mihaly Bencze
PP25514. If ak, bk, xk > 0 (k = 1, 2, ..., n) and x =n�
k=1
xk then�
n−2n−1
n�k=1
a2k +n�
k=1
a2kxk
x−xk
��n−2n−1
n�k=1
b2k +n�
k=1
b2kxkx−xk
�≥
≥ 4(n−1)2
��
1≤i<j≤n
�aibiajbj
�2
.
Mihaly Bencze
PP25515. If ak > 0, λk > 0 (k = 1, 2, ..., n) andn�
k=1
λk = 1 then
n
�n�
k=1
ak ≤ 1n
�aλ11 aλ2
2 ...aλnn + aλ2
1 aλ32 ...aλ1
n + ...+ aλn1 aλ1
2 ...aλn−1n
�≤ 1
n
n�k=1
ak.
326 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
(A generalization of Heinz-mean).
Mihaly Bencze
PP25516. In all scalene triangle ABC holds� a(b−a)
(b+c)(ma−mb)<
4(2s2−3Rr)3(s2+r2+2Rr)
.
Mihaly Bencze
PP25517. In all scalene triangle ABC holds8Rr(5s2+r2+4Rr)
3(s2+r2+2Rr)+� b3−a3
ma−mb<
2(9s2+r2+4Rr)3 .
Mihaly Bencze
PP25518. In all scalene triangle ABC holds� b2−a2
ma−mb< 16s
3 .
Mihaly Bencze
PP25519. Compute� � (x−sin y cos y−sin2 x)(y−sinx cosx−cos2 y)dxdy
(1−sin 2x)(1−sin 2y) .
Mihaly Bencze
PP25520. In all scalene triangle ABC holds94
� (b−a)mamb
m3a−m3
b≤� ma+mb
a+b ≤ 98
� (b−a)(m2a+m2
b)m3
a−m3b
.
Mihaly Bencze
PP25521. In all acute triangle ABC holds�
(1 + sinA) (1 + cosA) ≥ (3+2√2)sr
R2 .
Mihaly Bencze
PP25522. In all acute triangle ABC holds� 2+3 sinA+2 cos 2A+sinA cos 2A
1+cos 2A ≥ (17+12√2)s
R .
Mihaly Bencze
PP25523. In all acute triangle ABC holds��1 + 1
sinA
� �1 + 1
cosB
�≥ 3(3 + 2
√2).
Mihaly Bencze
Proposed Problems 327
PP25524. Compute� �
�
x+√
y2+y�
(y+√x2+x)dxdy
(x+√x2+x+
√x+
√x+1+1)
�
y+√
y2+y+√y+1+1
� .
Mihaly Bencze
PP25525. Let (an)n≥1 be an arithmetical progression formed with naturalnumbers. Prove that if an thermen of the given arithmetical progression is2025 then the given progression have infinitely many therms formed byperfect square.
Mihaly Bencze
PP25526. If xn+1 · xn = x2n + 1 for all n ≥ 1 and x1 > 1 then compute
limn→∞
n�√
e−�xn+1
xn
�n�.
Mihaly Bencze
PP25527. If xk > 1 (k = 1, 2, ..., n) , then
�cyclic
logx1
�x20142 + x20082 − x20072 + 1
�≥ 2011n.
Mihaly Bencze
PP25528. If xk > 1 (k = 1, 2, ..., n) , then
�cyclic
logx1
�x20142 + x20112 + 2x20082 − x20072 + 1
�≥ 2008n.
Mihaly Bencze
PP25529. If Ak ∈ Mn (C) (k = 1, 2, ..., n) such thatAn−1
1 = −A2A3...An;An−12 = −A3A4...AnA1, ..., A
n−1n = − (A1A2...An−1)
then An1 = An
2 = ... = Ann.
Mihaly Bencze
PP25530. Solve in R the equation
2�√
2x − 1 +√3x − 2x + ...+
�(n+ 1)x − nx
�= (n+ 1)x + n− 1 when
n ∈ N∗.
Mihaly Bencze
328 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25531. Let (an)n≥1 be an arithmetical progression formed by naturalnumbers. Prove that if an term of the given arithmetical progression is 2197the given progression have infinitely many therms formed by perfect cubes.
Mihaly Bencze
PP25532. If ak > 1 (k = 1, 2, ..., n) and
t =a21
(n−1)a2+a3+
a22(n−1)a3+a4
+ ...+ a2n(n−1)a1+a2
thenn�
k=1
logak t ≥ n.
Mihaly Bencze
PP25533. If xk > 1 (k = 1, 2, ..., n) , then�
cyclic
logx1
�2x32 − 1
�≥ n.
Mihaly Bencze
PP25534. If xk > 1 (k = 1, 2, ..., n) , then�cyclic
logx1
�x122 + x42 − x2 + 1
�≥ 9n.
Mihaly Bencze
PP25535. If xk > 1 (k =, 2, ..., n) , then�cyclic
logx1
�x122 − x92 + 2x42 − x2 + 1
�≥ 4n.
Mihaly Bencze
PP25536. If xk > 1 (k = 1, 2, ..., n) then�
cyclic
log2x31+x2
1
�2x42 + 1
�≥ n.
Mihaly Bencze
PP25537. If xk > 1 (k = 1, 2, ..., n) , then�cyclic
logx1
�2x42 − x32 − x22 + 1
�≥ 3n.
Mihaly Bencze
PP25538. If xk > 1 (k = 1, 2, ..., n) then�
cyclic
logx1
�2x42 − 2x32 + 1
�≥ 2n.
Mihaly Bencze
Proposed Problems 329
PP25539. If xk > 0 (k = 1, 2, ..., n) then�cyclic
logx1+1
�x42 + 9x32 + 31x22 + 49x2 + 31
�≥ 4n.
Mihaly Bencze
PP25540. If xk > 1 (k = 1, 2, ..., n) and m ∈ N∗
then�
cyclic
logx1
�x2m2 − (2m− 1)x2 + 2m− 1
�≥ n.
Mihaly Bencze
PP25541. If ak > 1 (k = 1, 2, ..., n) and λ > 0 then�cyclic
loga1�a3λ2 − 2a2λ2 + 4
�≥ 2nλ.
Mihaly Bencze
PP25542. If ak > 1 (k = 1, 2, ..., n) then�
cyclic
loga31+a1
�a42 + a22 + 1
�≥ n.
Mihaly Bencze
PP25543. In all triangle ABC holds� a(a+b)(a+c)
(b+c)(b2+c2+ab+ac)≤ 3
2 .
Mihaly Bencze
PP25544. In all triangle ABC holds� a(a+b)(a+c)
(b+c)(bc+(a+c)(a+b)) ≤ 2.
Mihaly Bencze
PP25545. In all triangle ABC holds� (b+c)√c+a−b
a√b
+� (c+a)
√b+c−a
b√a
≤ s2−r2−RrRr .
Mihaly Bencze
PP25546. In all triangle ABC holds� √
c(a+b−c)
a+b ≤ 32 .
Mihaly Bencze
PP25547. In all triangle ABC holds17s2+9r2+18Rr4(s2+r2+2Rr)
− 12
�s2−r2−Rrs2+r2+2Rr
�2≤
√2s� 1√
a+b≤ 4s2+2r2+5Rr
s2+r2+2Rr.
Mihaly Bencze
330 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25548. In all triangle ABC holds (b+c)√c+a−b
a√b
+ (c+a)√b+c−a
b√a
≤ (b+c)(c+a)ab
and his permutations.
Mihaly Bencze
PP25549. In all scalene triangle ABC holds���ctgA−B2 ctgC
2
��+���� sin2 A+sin2 B
sin(A−B) sinC
��� > 8.
Mihaly Bencze
PP25550. In all triangle ABC holds�s2 + r2 + 4Rr
� �s2 + r2 + 2Rr
�≥ 24Rr
�s2 − r2 −Rr
�.
Mihaly Bencze
PP25551. In all triangle ABC holds2(s2−r2−Rr)s2+r2+2Rr
≤ 2 + 3 3
�2Rr
s2+r2+2Rr.
Mihaly Bencze
PP25552. In all triangle ABC holds
4�
s2−r2−Rrs2+r2+2Rr
�2≤ 2 + 2
�s2+r2−Rrs2+r2+2Rr
�+ 3 3
��2Rr
s2+r2+2Rr
�2.
Mihaly Bencze
PP25553. In all triangle ABC holds��2
3
� ra3√
s2r ≤ 2.
Mihaly Bencze
PP25554. In all triangle ABC holds (4R+ r)3 + 15s2r ≥ 12s2R+ 6s23√s2r
Mihaly Bencze
PP25555. In all triangle ABC holds� (b+c)2(c+a)(a+b)
a2(c+a)(a+b)+bc(b+c)2≥ 2�s2+r2+2Rrs2−r2−Rr
�2.
Mihaly Bencze
PP25556. In all triangle ABC holds2(s2−r2−Rr)s2+r2+2Rr
≥ 136 − s2+r2+4Rr
2(s2−r2−4Rr).
Mihaly Bencze
Proposed Problems 331
PP25557. In all triangle ABC holds 3� 1
3a+b ≤ s2+r2+4Rr16sRr + 5s2+r2+4Rr
2s(s2+r2+2Rr).
Mihaly Bencze
PP25558. In all triangle ABC holds2(s2−r2−Rr)s2+r2+2Rr
+ s2+r2+4Rr2(s2−r2−4Rr)
≤ 52 .
Mihaly Bencze
PP25559. In all triangle ABC holds�
4
�a2
b ≥ 3 4
�2s3 .
Mihaly Bencze
PP25560. If x, y, z > 0 then�� x4
x3+y3
��� y3
x(x3+y3)
�≥ 9
4 .
Mihaly Bencze
PP25561. In all triangle ABC holds� a
b2+c2≥ 8(s2−r2−Rr)
5(s2+r2+2Rr).
Mihaly Bencze
PP25562. In scalene triangle ABC holds� 1
|a2−b2| ≥3s2−7r2−28Rrs2−r2−4Rr
.
Mihaly Bencze
PP25563. In all triangle ABC holds6(s2−r2−Rr)s2+r2+2Rr
≥ 92 + 7
8
�s2+r2−12Rrs2+r2+4Rr
�.
Mihaly Bencze
PP25564. In all triangle ABC holds� (b+c)(c+a)
(b+c)(c+a)+ab ≤ 5(s2+r2+2Rr)2(s2−r2−Rr)
.
Mihaly Bencze
PP25565. In all triangle ABC holds λ�
3�
a2b2s(s2+r2+4Rr)
− 1�≥ s2−7r2−10Rr
2(s2+r2+2Rr)
for all λ ∈�14 ,
12
�.
Mihaly Bencze
PP25566. In all triangle ABC holds 4�
s2−r2−Rrs2+r2+2Rr
�≥�2√2− 1
� �
a2b�
ab2+ 3.
Mihaly Bencze
332 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25567. In all triangle ABC holds2�s2 − r2 −Rr
�2 ≤�s2 + r2 + 2Rr
� �2s2 + 2r2 +Rr
�.
Mihaly Bencze
PP25568. In all triangle ABC holds��(c+ a)2 (b+ c)− ab (c+ a) + b2 (b+ c)
�≥ 4s2
�s2 + r2 + 2Rr
�2.
Mihaly Bencze
PP25569. In all triangle ABC holds
6�
s2−r2−Rrs2+r2+2Rr
�2−3
�
s2+r2−2Rrs2+r2+2Rr
�
2
�
(s2−r2−Rr)2+3R2r2
(s2+r2+2Rr)2− s2+r2−2Rr
s2+r2+2Rr
� ≥� (b+c)2
a2+(b+c)2.
Mihaly Bencze
PP25570. In all triangle ABC holds� a(a+c)(a+b)
(b+c)(bc+(c+a)(a+b)) ≤ 2.
Mihaly Bencze
PP25571. In all triangle ABC holds9(s2+r2+2Rr)5s2+r2+4Rr
≤ s2+r2+2Rr4s2
≤ 3 3√s2+r2+2Rr3√2Rr+ 3√s2+r2+2Rr
.
Mihaly Bencze
PP25572. In all convex quadrilateral ABCD holds��(b+ c)2 (c+ d)− a2b
�≥�
�(b+ c)3 − a3
�.
Mihaly Bencze
PP25573. In all triangle ABC holds1).� ma
2ma+mb+mc> 2
3
2).� cos A
2
2 cos A2+cos B
2+cos C
2
> 23
Mihaly Bencze
PP25574. In all triangle ABC holds�1 + 3
�2Rr
s2+r2+2Rr
�3≤ 4s2
s2+r2+2Rr≤�
5s2+r2+4Rr3(s2+r2+2Rr)
�3.
Mihaly Bencze
Proposed Problems 333
PP25575. If x, y, z, t ∈ R then1
sin2 x+sin2 y+sin2 z+sin2 t+ 1
cos2 x+cos2 y+cos2 z+cos2 t≥
≥ 12 + 64
27
�sin2 x sin2 y sin2 z sin2 t+ cos2 x cos2 y cos2 z cos2 t
�.
Mihaly Bencze
PP25576. In all tetrahedron ABCD holds
288r2 + 16 (ha + hb + hc + hc)
2 + 16 (ra + rb + rc + rd)
2 ≥
≥ 70
�1
1hahb
+ 1hahc
+ 1hahd
+ 1hbhc
+ 1hbhd
+ 1hchc
+ 11
rarb+ 1
rarc+ 1
rard+ 1
rbrc+ 1
rbrd+ 1
rcrc
�.
Mihaly Bencze
PP25577. Solve in Z the equationx21−x2
x3x4−x25+
x22−x3
x4x5−x26+ ...+ x2
n−x1
x2x3−x4= x2
x1+ x3
x2+ ...+ x1
xn.
Mihaly Bencze
PP25578. In all triangle ABC holds�
sin A2 sin B
2 ≤ 3 + r2R −
�6 + r
2R .
Mihaly Bencze
PP25579. In all triangle ABC holds
1).� (a+b)2
a+b−c ≥ 8s
2).� (ma+mb)
2
ma+mb−mc≥ 4 (ma +mb +mc)
3).� (cos A
2+cos B
2 )2
cos A2+cos B
2−cos C
2
≥� cos A2
4).� a2
3a−b−c ≥ 2s
Mihaly Bencze
PP25580. In all triangle ABC holds� w2
a(b+c−a)bc +
8ss2+r2+2Rr
=(5s2+r2+4Rr)
2
2s(s2+r2+2Rr)2.
Mihaly Bencze
PP25581. In all triangle ABC holds� bc
w2a≥ 4s.
Mihaly Bencze
334 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25582. In all triangle ABC holds(s2+r2+4Rr)
2
2(3s2−r2−4Rr)+� bw2
ac ≤ 2
�s2 − r2 − 4Rr
�.
Mihaly Bencze
PP25583. In all triangle ABC holds� (b+c−a)w2
abc ≥ 4s3
3s2−r2−4Rr.
Mihaly Bencze
PP25584. In all triangle ABC holds� (b+c−a)3w2
abc ≥ 4s(s2−2r2−8Rr)
2
3s2−r2−4Rr.
Mihaly Bencze
PP25585. In all triangle ABC holds� (b+c)w2
abc ≤ 3s.
Mihaly Bencze
PP25586. In all triangle ABC holds2(s2−r2−4Rr)
2
3s2−r2−4Rr+� a2w2
abc ≤ 2
�s2 − r2 − 4Rr
�.
Mihaly Bencze
PP25587. In all triangle ABC holds 2s2(2R−r)2
3s2−r2−4Rr+� w2
ar2a
bc ≤ (4R+ r)2−2s2.
Mihaly Bencze
PP25588. In all triangle ABC holds 43
�s2−r2−Rrs2+r2+2Rr
�2+� w2
abc ≤ 3.
Mihaly Bencze
PP25589. In all triangle ABC holds
4�
s2−r2−Rrs2+r2+2Rr
�2+� w2
abc = 3 + 4
�s2+r2−2Rrs2+r2+2Rr
�.
Mihaly Bencze
PP25590. In all triangle ABC holds1).� b+c
a2≥ s2+r2+4Rr
2sRr
2).� a
(s−a)2≥ 2(4R+r)
sr
3).� hb+hc
h2a
≥ 2r
4).� rb+rc
r2a≥ 2
r
5).� sin2 B
2+sin2 C
2
sin4 A2
≥ 2(s2+r2−8Rr)r2
Proposed Problems 335
6).� cos2 B
2+cos2 C
2
cos4 A2
≥ 2 + 2�4R+r
s
�2
Mihaly Bencze
PP25591. If xk > 0 (k = 1, 2, ..., n) thenn�
k=1
1xk(1+x2
k)≥ 8n5
�
n+n�
k=1
xk
�4 .
Mihaly Bencze
PP25592. Solve in Z the equation 2016x2 − 2015y2 = 1.
Mihaly Bencze
PP25593. Solve in N the equationn�
k=1
xxkk = yy.
Mihaly Bencze
PP25594. If x, y, z ∈ (0, 1) and x+ y + z = 1 then� xλ+1
1−xλ + 13λ−1
≥� (x+y)λ+1
2λ−(x+y)λfor all λ ≥ 1.
Mihaly Bencze
PP25595. If xk ∈ (0, 1) (k = 1, 2, ..., n) , then
n�k=1
kxλ+1k
1−xλ+1k
≥
�
n�
k=1xk
�λ+1
�
n(n+1)2
�λ−�
n�
k=1xk
�λ for all λ ≥ 1.
Mihaly Bencze
PP25596. Let ABC be a triangle and M ∈ Int (ABC). Prove that
1).�
MA2 · ra ≥ 14R+r
�a2rbrc
2).�
MA2 sin2 A2 ≥ 2R
2R−r
�a2 sin2 B
2 sin2 C2
3).�
MB2 cos2 A2 ≥ 2R
4R+r
�a2 cos2 B
2 cos2 C2
Mihaly Bencze
PP25597. Solve in Z the equation 5xy+z + 7
zt+u = 2.
Mihaly Bencze
336 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25598. If a > 1 and xk > 0 (k = 1, 2, ..., n) , then
n�k=1
(axk − 1) ≤�a
1n
n�
k=1xk
− 1
�n
.
Mihaly Bencze
PP25599. Let ABC be a triangle and M ∈ Int (ABC) . If α = MAB∡,
β = MBC∡, γ = MCA∡ then ctgα+ ctgβ + ctgγ ≥ s2−r(4R+r)2sr + 3 3
�2R2
sr .
Mihaly Bencze
PP25600. In all triangle ABC holdsu2λ(ab)2λ−1
(ha−r)(hb−r) +r2λ(bc)2λ−1
(hb−r)(hc−r) +t2λ(ca)2λ−1
(hc−r)(ha−r) ≥(uab+rbc+tca)2λ
32λ−2r2(5s2+r2+4Rr).
Mihaly Bencze
PP25601. If uk, vk > 0 (k = 1, 2, 3) then
u1
�v2v3v1
�lg v2v3 + u2
�v3v1v2
�lg v3v1 + u3
�v1v2v3
�lg v1v2 ≥ 3 3
√u1u2u3.
Mihaly Bencze
PP25602. In all triangle ABC holdsu2λa2λ−1
ha−r + v2λb2λ−1
hb−r + t2λc2λ−1
hc−r ≥ (ua+vb+tc)2λ
4·32λ−2sr.
Mihaly Bencze
PP25603. If a, b, c > 0, then determine all x, y ∈ R for which a, b, c are in
geometrical progression if and only if
������
ax (ab)y bx
bx (bc)y cx
cx (ca)y ax
������= 0.
Mihaly Bencze
PP25604. In all triangle ABC holds�
ra�bca
�lg bc ≥ 3
3√s2r.
Mihaly Bencze
PP25605. Let A1A2...An be a convex polygon. Prove that�n�
k=1
AksinAk
�sin
�1
(n−2)π
n�k=1
A2k
�≥ (n− 2)π.
Mihaly Bencze
Proposed Problems 337
PP25606. If xk > 0 (k = 1, 2, ..., n) thenn�
k=1
xk�0
e−t2dt ≤ narctg
�1n
n�k=1
xk
�.
Mihaly Bencze
PP25607. If zk ∈ C (k = 1, 2, ..., n) , thenn�
k=1
z2k ≥ Im (�
z1z2) .
Mihaly Bencze
PP25608. In all triangle ABC holds�� A
sinB
�sin�1π
�AB�≥ π.
Mihaly Bencze
PP25609. In all triangle ABC holds�� 1
ra sinA
�sin�r� A
ra
�≥ 1
r .
Mihaly Bencze
PP25610. In all triangle ABC holds
1).� 1
ha−r =2(s2−r2−Rr)r(s2+r2+2Rr)
2).� 1
(ha−r)(hb−r) =s2+r2−2Rr
r2(s2+r2+2Rr)
Mihaly Bencze
PP25611. In all triangle ABC holds� a2λ−1
ha−r ≥�23
�2λ−2 s2λ−1
r for allλ ∈ (−∞, 0] ∪ [1,+∞) .
Mihaly Bencze
PP25612. In all triangle ABC holds�
log ab+c
4bb+4(c+a) ≥
32 .
Mihaly Bencze
PP25613. Determine all function f : R → R such thatn�
k=1
xk ≤n�
k=1
5xkf (xk) ≤�
n�k=1
xk
�5
n�
k=1xk
for all xk ∈ R (k = 1, 2, ..., n) .
Mihaly Bencze
PP25614. Determine all functions N∗ → N∗ such thatn�
k=1
kf2 (k) = n2f2(n+1)4 for all n ∈ N∗.
Mihaly Bencze
338 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25615. Solve in R the following system:
�2−
√2 + x1 = x2�
2−√2 + x2 = x3
−−−−−−−−−�2−
√2 + xn = x1
..
Mihaly Bencze
PP25616. If xk ∈ R (k = 1, 2, ..., n) such that
x1√x2x3...xn = 1
2x2
√x3x4...xnx1 =
23
−−−−−−−−−−xn
√x1x2...xn−1 =
nn+1
thenn�
k=1
√xk
k2= n
(n+1)34.
Mihaly Bencze
PP25617. If a1 = 1 and an+1 = an + a[n+12 ] for all n ≥ 1 then compute
∞�n=1
1a2n.
Mihaly Bencze
PP25618. If xk ∈�ea, eb
�(k = 1, 2, ..., n) then�
n�k=1
1lnxk
�ln
�n�
k=1
xk
�≤ n2(a+b)2
4ab .
Mihaly Bencze
PP25619. Solve in R the following system
3|x1+1| = 2 + 3x2 + 2 |3x3 − 1|3|x2+1| = 2 + 3x3 + 2 |3x4 − 1|−−−−−−−−−−−−−3|xn+1| = 2 + 3x1 + 2 |3x2 − 1|
.
Mihaly Bencze
PP25620. Let f : N∗ → N∗ for which f (1) = 1, f (p) = 1 + f (p− 1) for all
p prime and f (p1p2...pn) =n�
k=1
f (pk) for all prime p1, p2, ..., pn. Prove that
n�k=1
2f(k) ≤�n(n+1)
2
�2≤
n�k=1
3f(k).
Mihaly Bencze
Proposed Problems 339
PP25621. Solve in R the following system:
�x21 + 4
�52x2−10x3+7 = x4�
x22 + 4�52x3−10x4+7 = x5
−−−−−−−−−−−�x2n + 4
�52x1−10x2+7 = x3
.
Mihaly Bencze
PP25622. If ak ∈ (0, 1) (k = 1, 2, ..., n) and λ > 0 then�
cyclic
loga12λa2a2+λ2 ≥ n
2 .
Mihaly Bencze
PP25623. Solve in R the following system:
5x1 = 3x2+{x3}
5x2 = 3x3+{x4}
−−−−−−5xn = 3x1+{x2}
, when {·}
denote the fractional part.
Mihaly Bencze
PP25624. Prove that 2n+1 + 3n+1−12·3n + 4n+1−1
3·4n + 6n+1−15 ≥ 4n+ 1 for all
n ∈ N.
Mihaly Bencze
PP25625. If xk ≥ 2 (k = 1, 2, ..., n) , then�
cyclic
(log2 a1)2
1+2 log2 a2≥ n
3 .
Mihaly Bencze
PP25626. Solve in C the dollowing system:
x1x2 + x3 + x4 = 2x5 + x26x2x3 + x4 + x5 = 2x6 + x27−−−−−−−−−−−xnx1 + x2 + x3 = 2x4 + x25
.
Mihaly Bencze
PP25627. If ak > 0 (k = 1, 2, ..., n) , thena1a2
+ 2�
a2a3
+ 3 3
�a3a4
+ ...+ n n
�ana1
≥ n(n+1)2 .
Mihaly Bencze
340 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25628. Determine all functions f : R → R such thatf (f (x) y) + f (xf (y)) = xy + f (xy) for all x, y ∈ R.
Mihaly Bencze
PP25629. If S =n�
k=1
zk when zk ∈ C (k = 1, 2, ..., n) and t ∈ N then
(n− 2t) |s|2 + t2n�
k=1
|zk|2 =n�
k=1
|s− tzk|2 .
Mihaly Bencze
PP25630. Solve in R the following system
2016x1 + 2017x2 = 2019x3 + 20142016x2 + 2017x3 = 2019x4 + 2014−−−−−−−−−−−−−−2016xn + 2017x1 = 2019x2 + 2014
.
Mihaly Bencze
PP25631. If xk > 0 (k = 1, 2, ..., n) then�
cyclic
�3�x21 + x22
�+ 10x1x2 ≤ 4
n�k=1
xk.
Mihaly Bencze
PP25632. Solve in R the following system
2sin 3x1 = sin3 x2 + 8sinx3
2sin 3x2 = sin3 x3 + 8sinx4
−−−−−−−−−−−2sin 3xn = sin3 x1 + 8sinx2
.
Mihaly Bencze
PP25633. Solve in R the following system:
22x1−1
= 1 + log2 (1 + x2)
22x2−1
= 1 + log2 (1 + x3)−−−−−−−−−−−−22
xn−1= 1 + log2 (1 + x1)
.
Mihaly Bencze
Proposed Problems 341
PP25634. If ak > 0 (k = 1, 2, ..., n) , then
2nn�
k=1
n√ak ≤ 4
n�k=1
√ak + 2n (n− 2) .
Mihaly Bencze
PP25635. Determine all increasing function f : R → R for whichf (xf (y) + yf (z) + zf (x)) = f (x) y + f (y) z + f (z)x for all x, y, z ∈ R.
Mihaly Bencze
PP25636. If zk ∈ C (k = 1, 2, ..., n) such that Re (zk) ≥ 0 (k = 1, 2, ..., n) ,
then 2n�
k=1
|zk|2 +�
(z1z2 + z1z2) ≥ 0.
Mihaly Bencze
PP25637. Determine all function f : (0,+∞) → R for which�n�
k=1
xk
�lg
�n�
k=1
xk
�≤�x2x3...xnf (x1) ≤ f
�n�
k=1
xk
�for all xk > 0
(k = 1, 2, ..., n) .
Mihaly Bencze
PP25638. If ε = −1+i√3
2 then solve on C the following system
(z1 − 1)2 + (z2 − ε)2 + (z3 − ε)2 = 3z24(z2 − 1)2 + (z3 − ε)2 + (z4 − ε)2 = 3z25−−−−−−−−−−−−−−−−(zn − 1)2 + (z1 − ε)2 + (z2 − ε)2 = 3z23
.
Mihaly Bencze
PP25639. Solve in R the following system
1 + log5 x1 = log2�4 +
√5x2�
1 + log5 x2 = log2�4 +
√5x3�
−−−−−−−−−−−−1 + log5 xn = log2
�4 +
√5x1�
.
Mihaly Bencze
342 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25640. If xk > 0 (k = 1, 2, ..., n) then
1).� (1+x3)
2
1+33√
1+(x1x2)2≥ n
2).� (1+x2)(1+x3)
1+33√
1+(x1x2)2≥ n
Mihaly Bencze
PP25641. If xk > 0 (k = 1, 2, ..., n) such that�
cyclic
n2x1
x2+x3 = n2 then
x1 = x2 = ... = xn.
Mihaly Bencze
PP25642. If A (a) , B (b) , C (c) then determine all x, y, z ∈ R for whichABC is equilateral if and only if ax (b− c) + by (c− a) + cz (a− b) = 0.
Mihaly Bencze
PP25643. Let be bk ≥ 1 (k = 1, 2, ..., n) be a geometrical progression with
ratio q. Prove that n lg b�lg b1 +
n−12 lg q
�≤
n�k=1
�lg Sk
k
�2≤
n (lg b1)2 + n(n−1)
2 lg b1 lg q +n(n−1)(2n−1)
12 (lg q)2 when Sn = b1 + b2 + ...+ bn.
Mihaly Bencze
PP25644. Solve in R the folowing system:
1 +�3 + 2
√2�x1 ≤ 6
�√2 + 1
�x2
1 +�3 + 2
√2�x2 ≤ 6
�√2 + 1
�x3
−−−−−−−−−−−−−1 +�3 + 2
√2�xn ≤ 6
�√2 + 1
�xn
.
Mihaly Bencze
PP25645. If ÷a1, a2, ..., an are an arithmetical progression with positive
numbers and Sn =n�
k=1
ak then
a1Sn ≤n�
k=1
�Skk
�2≤ na21 +
12a1r (n− 1) + 1
12n (n− 1) (2n− 1) r2 when r
denote the ratio.
Mihaly Bencze
Proposed Problems 343
PP25646. If ak > 0 (k = 1, 2, ..., n) such that�
cyclic
na1+a2+...+an−1
an = nn then
a1 = a2 = ... = an.
Mihaly Bencze
PP25647. If a, b ∈ R such that (a+ 1)2 + b2 ≥ 4 then�a3 − 3ab2
�2+�3a2b− b3
�2 ≥ 1.
Mihaly Bencze
PP25648. If ak ∈ (0, 1) (k = 1, 2, ..., n) , then�
cyclic
log a1+a2+...+an−1n−1
an ≥ n.
Mihaly Bencze
PP25649. Prove thatn�
k=1
1(k+1)
m√k≤ m
�1− 1
m√n+1
�for all n,m ∈ N∗.
Mihaly Bencze
PP25650. Solve in R the following system:
log9�1 + x21 + x32
�= 2 log4 x3
log9�1 + x22 + x33
�= 2 log4 x4
−−−−−−−−−−−−−log9�1 + x2n + x31
�= 2 log4 x2
.
Mihaly Bencze
PP25651. Determine all functions f : R → R for which2 (f ◦ f) (x+ y + z+) = x+ y + z + f (1− x) + f (1− y) + f (1− z) for allx, y, z ∈ R.
Mihaly Bencze
PP25652. Solve in C the following system
x21 +3�x62 + 7x33 =
�x44 + 8x5
x22 +3�x63 + 7x34 =
�x45 + 8x6
−−−−−−−−−−−−−x21 +
3�x62 + 7x33 =
�x44 + 8x5
Mihaly Bencze
344 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015
PP25653. If a, b > 0 then solve in R the following system:
a+√x1 − a2 = b+
√x2 − b2
a+√x2 − a2 = b+
√x3 − b2
−−−−−−−−−−−−−a+
√xn − a2 = b+
√x1 − b2
.
Mihaly Bencze
PP25654. Solve in R the following system
x1�3 +
√5�lg x2
+ 20 = 10�5 +
√5�lg x3
x2�3 +
√5�lg x3
+ 20 = 10�5 +
√5�lg x4
−−−−−−−−−−−−−−−−−xn�3 +
√5�lg x1
+ 20 = 10�5 +
√5�lg x2
.
Mihaly Bencze
PP25655. If 1 < a1 < a2 < ... < an are natural numbers, thenn�
k=1
log2
�1− 1
a2k
�≥ lg2
n+22(n+1) .
Mihaly Bencze
PP25656. Determine all z ∈ C such that
�|z − (2016 + i)| =
√2
|z − (1 + 2016i)| ≤√2
.
Mihaly Bencze
PP25657. Prove that for all n ∈ N the equation z4n+2 − z4n+1 + z + 1 = 0have a root z1 ∈ C\R for which |z1| = 1.
Mihaly Bencze
PP25658. Determine the integer part oflog2 2011 + log3 2012 + log4 2013 + log5 2014 + log6 2015 + log7 2016.
Mihaly Bencze
PP25659. Determine all ak > 0 (k = 1, 2, ..., n) for whichn�
k=1
k3(ak)!a2k
= (n+ 1)!− 1.
Mihaly Bencze