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Jozsef Wildt International Math Competition
2009 Jozsef Wildt International Math Competition
2009 Jozsef Wildt International Math Competition
Part of
Jozsef Wildt International Math Competition
Subcontests
(30)
W. 30
1
Hide problems
Prove this combinatorial equation
Prove that
∑
0
≤
i
<
j
≤
n
(
i
+
j
)
(
n
i
)
(
n
j
)
=
n
(
2
2
n
−
1
−
(
2
n
−
1
n
)
)
\sum \limits_{0\leq i<j\leq n}(i+j) {{n}\choose{i}}{{n}\choose{j}}=n\left (2^{2n-1}-{{2n-1}\choose{n}} \right )
0
≤
i
<
j
≤
n
∑
(
i
+
j
)
(
i
n
)
(
j
n
)
=
n
(
2
2
n
−
1
−
(
n
2
n
−
1
)
)
W. 29
1
Hide problems
Prove this inequality holds for all triangle
Prove that for all triangle
△
A
B
C
\triangle ABC
△
A
BC
holds the following inequality
∑
c
y
c
(
1
−
3
tan
A
2
+
3
tan
A
2
)
(
1
−
3
tan
B
2
+
3
tan
B
2
)
≥
3
\sum \limits_{cyc} \left (1-\sqrt{\sqrt{3}\tan \frac{A}{2}}+\sqrt{3}\tan \frac{A}{2}\right )\left (1-\sqrt{\sqrt{3}\tan \frac{B}{2}}+\sqrt{3}\tan \frac{B}{2}\right )\geq 3
cyc
∑
(
1
−
3
tan
2
A
+
3
tan
2
A
)
(
1
−
3
tan
2
B
+
3
tan
2
B
)
≥
3
W. 28
1
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Prove this problem on Banach space and jensen additive mapping
Let
θ
\theta
θ
and
p
(
p
<
1
)
p(p<1)
p
(
p
<
1
)
) be nonnegative real numbers.Suppose that
f
:
X
→
Y
f:X\to Y
f
:
X
→
Y
is mapping with
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
and
∣
∣
2
f
(
x
+
y
2
)
−
f
(
x
)
−
f
(
y
)
∣
∣
Y
≤
θ
(
∣
∣
x
∣
∣
X
p
+
∣
∣
y
∣
∣
X
p
)
\left |\left| 2f\left (\frac{x+y}{2}\right )-f(x)-f(y) \right |\right|_Y \leq \theta\left (\left |\left |x\right |\right |_X^p +\left |\left |y\right |\right |_X^p \right )
2
f
(
2
x
+
y
)
−
f
(
x
)
−
f
(
y
)
Y
≤
θ
(
∣
∣
x
∣
∣
X
p
+
∣
∣
y
∣
∣
X
p
)
for all
x
x
x
,
y
∈
Z
y\in \mathbb{Z}
y
∈
Z
with
x
⊥
y
x\perp y
x
⊥
y
where
X
X
X
is an orthogonality space and
Y
Y
Y
is a real Banach space.Prove that there exists a unique orthogonally Jensen additive mapping
T
:
X
→
Y
T:X\to Y
T
:
X
→
Y
, namely a mapping
T
T
T
that satisfies the so-called orthogonally Jensen additive functional equation
2
f
(
x
+
y
2
)
=
f
(
x
)
+
f
(
y
)
2f\left (\frac{x+y}{2}\right )=f(x)+f(y)
2
f
(
2
x
+
y
)
=
f
(
x
)
+
f
(
y
)
for all
x
x
x
,
y
∈
X
y\in \mathbb{X}
y
∈
X
with
x
⊥
y
x\perp y
x
⊥
y
, satisfying the property
∣
∣
f
(
x
)
−
T
(
x
)
∣
∣
Y
≤
2
p
θ
2
−
2
p
∣
∣
x
∣
∣
X
p
\left |\left|f(x)-T(x) \right |\right|_Y \leq \frac{2^p\theta}{2-2^p}\left |\left |x\right |\right |_X^p
∣
∣
f
(
x
)
−
T
(
x
)
∣
∣
Y
≤
2
−
2
p
2
p
θ
∣
∣
x
∣
∣
X
p
for all
x
∈
X
x\in X
x
∈
X
W. 27
1
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Prove this inequality on perfect number
Let
a
a
a
,
n
n
n
be positive integers such that
a
n
a^n
a
n
is a perfect number. Prove that
a
n
μ
>
μ
2
a^{\frac{n}{\mu}}> \frac{\mu}{2}
a
μ
n
>
2
μ
where
μ
\mu
μ
denotes the number of distinct prime divisors of
a
n
a^n
a
n
W. 26
1
Hide problems
Prove this hard inequality
If
a
i
>
0
a_i >0
a
i
>
0
(
i
=
1
,
2
,
⋯
,
n
i=1, 2, \cdots , n
i
=
1
,
2
,
⋯
,
n
) and
∑
i
=
1
n
a
i
k
=
1
\sum \limits_{i=1}^n a_i^k=1
i
=
1
∑
n
a
i
k
=
1
, where
1
≤
k
≤
n
+
1
1\leq k\leq n+1
1
≤
k
≤
n
+
1
, then
∑
i
=
1
n
a
i
+
1
∏
i
=
1
n
a
i
≥
n
1
−
1
k
+
n
n
k
\sum \limits_{i=1}^n a_i + \frac{1}{\prod \limits_{i=1}^n a_i} \geq n^{1-\frac{1}{k}}+n^{\frac{n}{k}}
i
=
1
∑
n
a
i
+
i
=
1
∏
n
a
i
1
≥
n
1
−
k
1
+
n
k
n
W. 25
1
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Prove this inequality on cyclic quadrilateral
Let
A
B
C
D
ABCD
A
BC
D
be a quadrilateral in which
A
^
=
C
^
=
9
0
∘
\widehat{A}=\widehat{C}=90^{\circ}
A
=
C
=
9
0
∘
. Prove that
1
B
D
(
A
B
+
B
C
+
C
D
+
D
A
)
+
B
D
2
(
1
A
B
⋅
A
D
+
1
C
B
⋅
C
D
)
≥
2
(
2
+
2
)
\frac{1}{BD}(AB+BC+CD+DA)+BD^2\left (\frac{1}{AB\cdot AD}+\frac{1}{CB\cdot CD}\right )\geq 2\left (2+\sqrt{2}\right )
B
D
1
(
A
B
+
BC
+
C
D
+
D
A
)
+
B
D
2
(
A
B
⋅
A
D
1
+
CB
⋅
C
D
1
)
≥
2
(
2
+
2
)
w. 24
1
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Prove this inequality for any point in the plane of triangle
If
K
K
K
,
L
L
L
,
M
M
M
denote the midpoints of the sides
A
B
AB
A
B
,
B
C
BC
BC
,
C
A
CA
C
A
in triangle
△
A
B
C
\triangle ABC
△
A
BC
, then for all
P
P
P
in the plane of triangle
△
A
B
C
\triangle ABC
△
A
BC
, we have
A
B
P
K
+
B
C
P
L
+
C
A
P
M
≥
A
B
⋅
B
C
⋅
C
A
4
⋅
P
K
⋅
P
L
⋅
P
M
\frac{AB}{PK}+\frac{BC}{PL}+\frac{CA}{PM} \geq \frac{AB\cdot BC \cdot CA}{4\cdot PK\cdot PL\cdot PM}
P
K
A
B
+
P
L
BC
+
PM
C
A
≥
4
⋅
P
K
⋅
P
L
⋅
PM
A
B
⋅
BC
⋅
C
A
W. 23
1
Hide problems
Prove this inequality
If
x
k
∈
R
x_k \in \mathbb{R}
x
k
∈
R
(
k
=
1
,
2
,
⋯
,
n
k=1, 2, \cdots , n
k
=
1
,
2
,
⋯
,
n
) and
m
∈
N
m \in \mathbb{N}
m
∈
N
then[*]
∑
c
y
c
(
x
1
2
−
x
1
x
2
+
x
2
2
)
m
≤
3
m
∑
k
=
1
n
x
k
2
m
\sum \limits_{cyc} \left (x_1^2 -x_1x_2+x_2^2 \right )^m \leq 3^m \sum \limits_{k=1}^n x_k^{2m}
cyc
∑
(
x
1
2
−
x
1
x
2
+
x
2
2
)
m
≤
3
m
k
=
1
∑
n
x
k
2
m
[*]
∏
c
y
c
(
x
1
2
−
x
1
x
2
+
x
2
2
)
m
≤
(
3
m
n
)
m
(
∑
k
=
1
n
x
k
2
m
)
n
\prod \limits_{cyc} \left (x_1^2 -x_1x_2+x_2^2 \right )^m \leq \left (\frac{3^m}{n}\right )^m \left (\sum \limits_{k=1}^n x_k^{2m}\right )^n
cyc
∏
(
x
1
2
−
x
1
x
2
+
x
2
2
)
m
≤
(
n
3
m
)
m
(
k
=
1
∑
n
x
k
2
m
)
n
W. 22
1
Hide problems
Prove this inequality
If
a
i
>
0
a_i >0
a
i
>
0
(
i
=
1
,
2
,
⋯
,
n
i=1, 2, \cdots , n
i
=
1
,
2
,
⋯
,
n
), then
(
a
1
a
2
)
k
+
(
a
2
a
3
)
k
+
⋯
+
(
a
n
a
1
)
k
≥
a
1
a
2
+
a
2
a
3
+
⋯
+
a
n
a
1
\left (\frac{a_1}{a_2} \right )^k + \left (\frac{a_2}{a_3} \right )^k + \cdots + \left (\frac{a_n}{a_1} \right )^k \geq \frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots + \frac{a_n}{a_1}
(
a
2
a
1
)
k
+
(
a
3
a
2
)
k
+
⋯
+
(
a
1
a
n
)
k
≥
a
2
a
1
+
a
3
a
2
+
⋯
+
a
1
a
n
for all
k
∈
N
k\in \mathbb{N}
k
∈
N
W. 21
1
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Prove this statement about riemann zeta function
If
ζ
\zeta
ζ
denote the Riemann Zeta Function, and
s
>
1
s>1
s
>
1
then
∑
k
=
1
∞
1
1
+
k
s
≥
ζ
(
s
)
1
+
ζ
(
s
)
\sum \limits_{k=1}^{\infty} \frac{1}{1+k^s}\geq \frac{\zeta (s)}{1+\zeta (s)}
k
=
1
∑
∞
1
+
k
s
1
≥
1
+
ζ
(
s
)
ζ
(
s
)
W. 20
1
Hide problems
Prove this trigonometric equation
If
x
∈
R
\
{
k
π
2
∣
k
∈
Z
}
x \in \mathbb{R}\backslash \left \{\frac{k\pi}{2}\ |\ k\in \mathbb{Z} \right \}
x
∈
R
\
{
2
kπ
∣
k
∈
Z
}
, then
(
∑
0
≤
j
<
k
≤
n
sin
(
2
(
j
+
k
)
x
)
)
2
+
(
∑
0
≤
j
<
k
≤
n
cos
(
2
(
j
+
k
)
x
)
)
2
=
sin
2
n
x
sin
2
(
n
+
1
)
x
sin
2
x
sin
2
2
x
\left (\sum \limits_{0\leq j<k\leq n} \sin (2(j+k)x)\right )^2 + \left (\sum \limits_{0\leq j<k\leq n} \cos (2(j+k)x)\right )^2 = \frac{\sin ^2 nx \sin ^2 (n+1)x}{\sin ^2x \sin^22x}
0
≤
j
<
k
≤
n
∑
sin
(
2
(
j
+
k
)
x
)
2
+
0
≤
j
<
k
≤
n
∑
cos
(
2
(
j
+
k
)
x
)
2
=
sin
2
x
sin
2
2
x
sin
2
n
x
sin
2
(
n
+
1
)
x
W. 19
1
Hide problems
Prove this inequality
If
x
k
>
0
x_k >0
x
k
>
0
(
k
=
1
,
2
,
⋯
,
n
k=1, 2, \cdots , n
k
=
1
,
2
,
⋯
,
n
), then
∑
k
=
1
n
(
x
k
1
+
x
1
2
+
x
2
2
+
⋯
+
x
k
2
)
2
≤
∑
k
=
1
n
x
k
2
1
+
∑
k
=
1
n
x
k
2
\sum \limits_{k=1}^n \left ( \frac{x_k}{1+x_1^2+x_2^2+\cdots +x_k^2} \right )^2 \leq \frac{\sum \limits_{k=1}^n x_k^2}{1+\sum \limits_{k=1}^n x_k^2}
k
=
1
∑
n
(
1
+
x
1
2
+
x
2
2
+
⋯
+
x
k
2
x
k
)
2
≤
1
+
k
=
1
∑
n
x
k
2
k
=
1
∑
n
x
k
2
W. 18
1
Hide problems
prove this cyclic inequality
If
a
a
a
,
b
b
b
,
c
>
0
c>0
c
>
0
and
a
b
c
=
1
abc=1
ab
c
=
1
, then
∑
c
y
c
a
+
b
+
c
n
a
2
n
+
3
+
b
2
n
+
3
+
a
b
≤
a
n
+
1
+
b
n
+
1
+
c
n
+
1
\sum \limits^{cyc} \frac{a+b+c^n}{a^{2n+3}+b^{2n+3}+ab} \leq a^{n+1}+b^{n+1}+c^{n+1}
∑
cyc
a
2
n
+
3
+
b
2
n
+
3
+
ab
a
+
b
+
c
n
≤
a
n
+
1
+
b
n
+
1
+
c
n
+
1
for all
n
∈
N
n\in \mathbb{N}
n
∈
N
W. 17
1
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Prove this inequality on two functions
If
a
a
a
,
b
b
b
,
c
>
0
c>0
c
>
0
and
a
b
c
=
1
abc=1
ab
c
=
1
,
α
=
m
a
x
{
a
,
b
,
c
}
\alpha = max\{a,b,c\}
α
=
ma
x
{
a
,
b
,
c
}
;
f
,
g
:
(
0
,
+
∞
)
→
R
f,g : (0, +\infty )\to \mathbb{R}
f
,
g
:
(
0
,
+
∞
)
→
R
, where
f
(
x
)
=
2
(
x
+
1
)
2
x
f(x)=\frac{2(x+1)^2}{x}
f
(
x
)
=
x
2
(
x
+
1
)
2
and
g
(
x
)
=
(
x
+
1
)
(
1
x
+
1
)
2
g(x)= (x+1)\left (\frac{1}{\sqrt{x}}+1\right )^2
g
(
x
)
=
(
x
+
1
)
(
x
1
+
1
)
2
, then
(
a
+
1
)
(
b
+
1
)
(
c
+
1
)
≥
m
i
n
{
{
f
(
x
)
,
g
(
x
)
}
∣
x
∈
{
a
,
b
,
c
}
\
{
α
}
}
(a+1)(b+1)(c+1)\geq min\{ \{f(x),g(x) \}\ |\ x\in\{a,b,c\} \backslash \{ \alpha \}\}
(
a
+
1
)
(
b
+
1
)
(
c
+
1
)
≥
min
{{
f
(
x
)
,
g
(
x
)}
∣
x
∈
{
a
,
b
,
c
}
\
{
α
}}
W. 16
1
Hide problems
Prove this inequality on number of divisors of a number
Prove that
∑
k
=
1
n
1
d
(
k
)
>
n
+
1
−
1
\sum \limits_{k=1}^n \frac{1}{d(k)}>\sqrt{n+1}-1
k
=
1
∑
n
d
(
k
)
1
>
n
+
1
−
1
For every
n
≥
1
n\geq 1
n
≥
1
,
d
(
n
)
d(n)
d
(
n
)
is the number of divisors of
n
n
n
W. 15
1
Hide problems
Prove this hard inequality on triangles
Let a triangle
△
A
B
C
\triangle ABC
△
A
BC
and the real numbers
x
x
x
,
y
y
y
,
z
>
0
z>0
z
>
0
. Prove that
x
n
cos
A
2
+
y
n
cos
B
2
+
z
n
cos
C
2
≥
(
y
z
)
n
2
sin
A
+
(
z
x
)
n
2
sin
B
+
(
x
y
)
n
2
sin
C
x^n\cos\frac{A}{2}+y^n\cos\frac{B}{2}+z^n\cos\frac{C}{2}\geq (yz)^{\frac{n}{2}}\sin A +(zx)^{\frac{n}{2}}\sin B +(xy)^{\frac{n}{2}}\sin C
x
n
cos
2
A
+
y
n
cos
2
B
+
z
n
cos
2
C
≥
(
yz
)
2
n
sin
A
+
(
z
x
)
2
n
sin
B
+
(
x
y
)
2
n
sin
C
W. 14
1
Hide problems
Prove this integral inequality
If the function
f
:
[
0
,
1
]
→
(
0.
+
∞
)
f:[0,1]\to (0.+\infty)
f
:
[
0
,
1
]
→
(
0.
+
∞
)
is increasing and continuous, then for every
a
≥
0
a\geq 0
a
≥
0
the following inequality holds:
∫
0
1
x
a
+
1
f
(
x
)
d
x
≤
a
+
1
a
+
2
∫
0
1
x
a
f
(
x
)
d
x
\int \limits_0^1 \frac{x^{a+1}}{f(x)}dx \leq \frac{a+1}{a+2} \int \limits_0^1 \frac{x^{a}}{f(x)}dx
0
∫
1
f
(
x
)
x
a
+
1
d
x
≤
a
+
2
a
+
1
0
∫
1
f
(
x
)
x
a
d
x
W. 13
1
Hide problems
Prove this inequality
If
a
k
>
0
a_k >0
a
k
>
0
[
k
=
k=
k
=
1, 2,
⋯
\cdots
⋯
,
n
n
n
], then prove the following inequality
(
∑
k
=
1
n
a
k
5
)
4
≥
1
n
(
2
n
−
1
)
5
(
∑
1
≤
i
<
j
≤
n
a
i
2
a
j
2
)
5
\left (\sum \limits_{k=1}^n a_k^5 \right )^4 \geq \frac{1}{n} \left (\frac{2}{n-1} \right )^5 \left (\sum \limits_{1\leq i<j\leq n} a_i^2a_j^2 \right )^5
(
k
=
1
∑
n
a
k
5
)
4
≥
n
1
(
n
−
1
2
)
5
(
1
≤
i
<
j
≤
n
∑
a
i
2
a
j
2
)
5
W. 12
1
Hide problems
Find all functions
Find all functions
f
:
(
0
,
+
∞
)
∩
Q
→
(
0
,
+
∞
)
∩
Q
f: (0, +\infty)\cap\mathbb{Q}\to (0, +\infty)\cap\mathbb{Q}
f
:
(
0
,
+
∞
)
∩
Q
→
(
0
,
+
∞
)
∩
Q
satisfying thefollowing conditions:[*]
f
(
a
x
)
≤
(
f
(
x
)
)
a
f(ax) \leq (f(x))^a
f
(
a
x
)
≤
(
f
(
x
)
)
a
, for every
x
∈
(
0
,
+
∞
)
∩
Q
x\in (0, +\infty)\cap\mathbb{Q}
x
∈
(
0
,
+
∞
)
∩
Q
and
a
∈
(
0
,
1
)
∩
Q
a \in (0, 1)\cap\mathbb{Q}
a
∈
(
0
,
1
)
∩
Q
[*]
f
(
x
+
y
)
≤
f
(
x
)
f
(
y
)
f(x+y) \leq f(x)f(y)
f
(
x
+
y
)
≤
f
(
x
)
f
(
y
)
, for every
x
,
y
∈
(
0
,
+
∞
)
∩
Q
x,y\in (0, +\infty)\cap\mathbb{Q}
x
,
y
∈
(
0
,
+
∞
)
∩
Q
W. 11
1
Hide problems
Find all real numbers m
Find all real numbers
m
m
m
such that
1
−
m
2
m
∈
{
x
∣
m
2
x
4
+
3
m
x
3
+
2
x
2
+
x
=
1
∀
x
∈
R
}
\frac{1-m}{2m} \in \{x\ |\ m^2x^4+3mx^3+2x^2+x=1\ \forall \ x\in \mathbb{R} \}
2
m
1
−
m
∈
{
x
∣
m
2
x
4
+
3
m
x
3
+
2
x
2
+
x
=
1
∀
x
∈
R
}
W. 10
1
Hide problems
Good functional and set problem
Let consider the following function set
F
=
{
f
∣
f
:
{
1
,
2
,
⋯
,
n
}
→
{
1
,
2
,
⋯
,
n
}
}
F=\{f\ |\ f:\{1,\ 2,\ \cdots,\ n\}\to \{1,\ 2,\ \cdots,\ n\} \}
F
=
{
f
∣
f
:
{
1
,
2
,
⋯
,
n
}
→
{
1
,
2
,
⋯
,
n
}}
[*] Find
∣
F
∣
|F|
∣
F
∣
[*] For
n
=
2
k
n=2k
n
=
2
k
prove that
∣
F
∣
<
e
(
4
k
)
k
|F|< e{(4k)}^{k}
∣
F
∣
<
e
(
4
k
)
k
[*] Find
n
n
n
, if
∣
F
∣
=
540
|F|=540
∣
F
∣
=
540
and
n
=
2
k
n=2k
n
=
2
k
W. 9
1
Hide problems
Find a real set on which this series is convergent
Let the series
s
(
n
,
x
)
=
∑
k
=
0
n
(
1
−
x
)
(
1
−
2
x
)
(
1
−
3
x
)
⋯
(
1
−
n
x
)
n
!
s(n,x)=\sum \limits_{k= 0}^n \frac{(1-x)(1-2x)(1-3x)\cdots(1-nx)}{n!}
s
(
n
,
x
)
=
k
=
0
∑
n
n
!
(
1
−
x
)
(
1
−
2
x
)
(
1
−
3
x
)
⋯
(
1
−
n
x
)
Find a real set on which this series is convergent, and then compute its sum. Find also
lim
(
n
,
x
)
→
(
∞
,
0
)
s
(
n
,
x
)
\lim \limits_{(n,x)\to (\infty ,0)} s(n,x)
(
n
,
x
)
→
(
∞
,
0
)
lim
s
(
n
,
x
)
W. 8
1
Hide problems
Prove this combinatorial formula
If
n
,
p
,
q
∈
N
,
p
<
q
n,p,q \in \mathbb{N}, p<q
n
,
p
,
q
∈
N
,
p
<
q
then
(
(
p
+
q
)
n
n
)
∑
k
=
0
n
(
−
1
)
k
(
n
k
)
(
(
p
+
q
−
1
)
n
p
n
−
k
)
=
(
(
p
+
q
)
n
p
n
)
∑
k
=
0
[
n
2
]
(
−
1
)
k
(
p
n
k
)
(
(
q
−
p
)
n
n
−
2
k
)
{{(p+q)n}\choose{n}} \sum \limits_{k=0}^n (-1)^k {{n}\choose{k}} {{(p+q-1)n}\choose{pn-k}}= {{(p+q)n}\choose{pn}} \sum \limits_{k=0}^{\left [\frac{n}{2} \right ]} (-1)^k {{pn}\choose{k}} {{(q-p)n}\choose{n-2k}}
(
n
(
p
+
q
)
n
)
k
=
0
∑
n
(
−
1
)
k
(
k
n
)
(
p
n
−
k
(
p
+
q
−
1
)
n
)
=
(
p
n
(
p
+
q
)
n
)
k
=
0
∑
[
2
n
]
(
−
1
)
k
(
k
p
n
)
(
n
−
2
k
(
q
−
p
)
n
)
W. 7
1
Hide problems
Prove this integration inequality
If
0
<
a
<
b
0<a<b
0
<
a
<
b
then
∫
a
b
(
x
2
−
(
a
+
b
2
)
2
)
ln
x
a
ln
x
b
(
x
2
+
a
2
)
(
x
2
+
b
2
)
d
x
>
0
\int \limits_a^b \frac{\left (x^2-\left (\frac{a+b}{2} \right )^2\right )\ln \frac{x}{a} \ln \frac{x}{b}}{(x^2+a^2)(x^2+b^2)} dx > 0
a
∫
b
(
x
2
+
a
2
)
(
x
2
+
b
2
)
(
x
2
−
(
2
a
+
b
)
2
)
ln
a
x
ln
b
x
d
x
>
0
W. 6
1
Hide problems
Problem related to partitions
Prove that
p
(
n
)
=
2
+
(
p
(
1
)
+
⋯
+
p
(
[
n
2
]
+
χ
1
(
n
)
)
+
(
p
2
′
(
n
)
+
⋯
+
p
[
n
2
]
−
1
′
(
n
)
)
)
p (n)= 2+ \left (p (1) + \cdots + p\left ( \left [\frac {n}{2} \right ] + \chi_1 (n)\right ) + \left (p'_2(n) + \cdots + p' _{ \left [\frac {n}{2} \right ] - 1}(n)\right )\right )
p
(
n
)
=
2
+
(
p
(
1
)
+
⋯
+
p
(
[
2
n
]
+
χ
1
(
n
)
)
+
(
p
2
′
(
n
)
+
⋯
+
p
[
2
n
]
−
1
′
(
n
)
)
)
for every
n
∈
N
n \in \mathbb {N}
n
∈
N
with
n
>
2
n>2
n
>
2
where
χ
\chi
χ
denotes the principal character Dirichlet modulo 2, i.e.
χ
1
(
n
)
=
{
1
if
(
n
,
2
)
=
1
0
if
(
n
,
2
)
>
1
\chi _1 (n) = \begin{cases} 1 & \text{if } (n,2)=1 \\ 0 &\text{if } (n,2)>1 \end{cases}
χ
1
(
n
)
=
{
1
0
if
(
n
,
2
)
=
1
if
(
n
,
2
)
>
1
with
p
(
n
)
p (n)
p
(
n
)
we denote number of possible partitions of
n
n
n
and
p
m
′
(
n
)
p' _m(n)
p
m
′
(
n
)
we denote the number of partitions of
n
n
n
in exactly
m
m
m
sumands.
W. 5
1
Hide problems
Prove that this does not have any integer solutions
Let
p
1
p_1
p
1
,
p
2
p_2
p
2
be two odd prime numbers and
α
\alpha
α
,
n
n
n
be positive integers with
α
>
1
\alpha >1
α
>
1
,
n
>
1
n>1
n
>
1
. Prove that if the equation
(
p
2
−
1
2
)
p
1
+
(
p
2
+
1
2
)
p
1
=
α
n
\left (\frac{p_2 -1}{2} \right )^{p_1} + \left (\frac{p_2 +1}{2} \right )^{p_1} = \alpha^n
(
2
p
2
−
1
)
p
1
+
(
2
p
2
+
1
)
p
1
=
α
n
does not have integer solutions for both
p
1
=
p
2
p_1 =p_2
p
1
=
p
2
and
p
1
≠
p
2
p_1 \neq p_2
p
1
=
p
2
.
W. 4
1
Hide problems
Problem on euler's totient fumction
Let
Φ
\Phi
Φ
denote the Euler totient function. Prove that for infinitely many
k
k
k
we have
Φ
(
2
k
+
1
)
<
2
k
−
1
\Phi (2^k+1) < 2^{k-1}
Φ
(
2
k
+
1
)
<
2
k
−
1
and that for infinitely many
m
m
m
one has
Φ
(
2
m
+
1
)
>
2
m
−
1
\Phi (2^m+1) > 2^{m-1}
Φ
(
2
m
+
1
)
>
2
m
−
1
W. 3
1
Hide problems
Find all such numbers
Let
Φ
\Phi
Φ
and
Ψ
\Psi
Ψ
denote the Euler totient and Dedekind‘s totient respectively. Determine all
n
n
n
such that
Φ
(
n
)
\Phi(n)
Φ
(
n
)
divides
n
+
Ψ
(
n
)
n +\Psi (n)
n
+
Ψ
(
n
)
.
W. 2
1
Hide problems
Find the area of the set
Find the area of the set
A
=
{
(
x
,
y
)
∣
1
≤
x
≤
e
,
0
≤
y
≤
f
(
x
)
}
A = \{(x, y)\ |\ 1 \leq x \leq e,\ 0 \leq y \leq f (x)\}
A
=
{(
x
,
y
)
∣
1
≤
x
≤
e
,
0
≤
y
≤
f
(
x
)}
, where \begin{tabular}{ c| c c c c |} &1 & 1& 1 & 1\\
f
(
x
)
f(x)
f
(
x
)
=&
ln
x
\ln x
ln
x
& 2
ln
x
\ln x
ln
x
& 3
ln
x
\ln x
ln
x
& 4
ln
x
\ln x
ln
x
\\ &
(
ln
x
)
2
{(\ln x)}^2
(
ln
x
)
2
&
4
(
ln
x
)
2
4{(\ln x)}^2
4
(
ln
x
)
2
&
9
(
ln
x
)
2
9{(\ln x)}^2
9
(
ln
x
)
2
&
16
(
ln
x
)
2
16{(\ln x)}^2
16
(
ln
x
)
2
\\ &
(
ln
x
)
3
{(\ln x)}^3
(
ln
x
)
3
&
8
(
ln
x
)
3
8{(\ln x)}^3
8
(
ln
x
)
3
&
27
(
ln
x
)
3
27{(\ln x)}^3
27
(
ln
x
)
3
&
64
(
ln
x
)
3
64{(\ln x)}^3
64
(
ln
x
)
3
\end{tabular}
W. 1
1
Hide problems
Hard Inequality
Let
a
a
a
,
b
b
b
,
c
c
c
be positive real numbers such that
a
+
b
+
c
=
1
a + b + c = 1
a
+
b
+
c
=
1
. Prove that
(
1
+
a
b
+
c
)
1
−
a
b
c
(
1
+
b
c
+
a
)
1
−
b
c
a
(
1
+
c
a
+
b
)
1
−
c
a
b
3
≥
64
\sqrt[3]{\left (\frac{1+a}{b+c}\right )^{\frac{1-a}{bc}}\left (\frac{1+b}{c+a}\right )^{\frac{1-b}{ca}}\left (\frac{1+c}{a+b}\right )^{\frac{1-c}{ab}}} \geq 64
3
(
b
+
c
1
+
a
)
b
c
1
−
a
(
c
+
a
1
+
b
)
c
a
1
−
b
(
a
+
b
1
+
c
)
ab
1
−
c
≥
64