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Contests
National and Regional Contests
Thailand Contests
Mathcenter Contest
2011 Mathcenter Contest + Longlist
2011 Mathcenter Contest + Longlist
Part of
Mathcenter Contest
Subcontests
(13)
11
1
Hide problems
2(ab+bc+ca)<= 5+ abc for a,b,c>0 with a+b+c=3
Let
a
,
b
,
c
∈
R
+
a,b,c\in R^+
a
,
b
,
c
∈
R
+
with
a
+
b
+
c
=
3
a+b+c=3
a
+
b
+
c
=
3
. Prove that
2
(
a
b
+
b
c
+
c
a
)
≤
5
+
a
b
c
2(ab+bc+ca)\le 5+ abc
2
(
ab
+
b
c
+
c
a
)
≤
5
+
ab
c
(Real Matrik)
10
1
Hide problems
sum 1/(1-p)^2 >=1 if pqr=1
Let
p
,
q
,
r
∈
R
p,q,r\in R
p
,
q
,
r
∈
R
with
p
q
r
=
1
pqr=1
pq
r
=
1
. Prove that
(
1
1
−
p
)
2
+
(
1
1
−
q
)
2
+
(
1
1
−
r
)
2
≥
1
\left(\frac{1}{1-p}\right)^2+\left(\frac{1}{1-q}\right)^2+\left(\frac{1}{1-r}\right)^2\ge 1
(
1
−
p
1
)
2
+
(
1
−
q
1
)
2
+
(
1
−
r
1
)
2
≥
1
(Real Matrik)
7
1
Hide problems
sum k_1=2n-3, sum 1/k_i=3
Given
k
1
,
k
2
,
.
.
.
,
k
n
∈
R
+
k_1,k_2,...,k_n\in R^+
k
1
,
k
2
,
...
,
k
n
∈
R
+
, find all the naturals
n
n
n
such that
k
1
+
k
2
+
.
.
.
+
k
n
=
2
n
−
3
k_1+k_2+...+k_n=2n-3
k
1
+
k
2
+
...
+
k
n
=
2
n
−
3
1
k
1
+
1
k
2
+
.
.
.
+
1
k
n
=
3
\frac{1}{k_1}+\frac{1}{k_2}+...+\frac{1}{k_n}=3
k
1
1
+
k
2
1
+
...
+
k
n
1
=
3
(Zhuge Liang)
5
1
Hide problems
sum a^3b^3/(a+b) >=1/2 (1/a+1/b+1/c) for a,,c>0 with abc=1
Let
a
,
b
,
c
∈
R
+
a,b,c\in R^+
a
,
b
,
c
∈
R
+
with
a
b
c
=
1
abc=1
ab
c
=
1
. Prove that
a
3
b
3
a
+
b
+
b
3
c
3
b
+
c
+
c
3
c
3
c
+
a
≥
1
2
(
1
a
+
1
b
+
1
c
)
\frac{a^3b^3}{a+b}+\frac{b^3c^3}{b+c}+\frac{c^3c^3}{c+a} \ge \frac12 \left(\frac{1}{a}+ \frac{1}{b}+\frac{1}{c}\right)
a
+
b
a
3
b
3
+
b
+
c
b
3
c
3
+
c
+
a
c
3
c
3
≥
2
1
(
a
1
+
b
1
+
c
1
)
(Zhuge Liang)
9 sl13
1
Hide problems
sum 1/\sqrt{2a+1} <=( 3/2)^{3/2} if a,b,c>0 with 3=a+b+c<= 3abc
Let
a
,
b
,
c
∈
R
+
a,b,c\in\mathbb{R^+}
a
,
b
,
c
∈
R
+
If
3
=
a
+
b
+
c
≤
3
a
b
c
3=a+b+c\le 3abc
3
=
a
+
b
+
c
≤
3
ab
c
, prove that
1
2
a
+
1
+
1
2
b
+
1
+
1
2
c
+
1
≤
(
3
2
)
3
/
2
\frac{1}{\sqrt{2a+1}}+ \frac{1}{\sqrt{2b+1}}+\frac{1}{\sqrt{2c+1}}\le \left( \frac32\right)^{3/2}
2
a
+
1
1
+
2
b
+
1
1
+
2
c
+
1
1
≤
(
2
3
)
3/2
(Real Matrik)
8 sl12
1
Hide problems
sum a^{11}/b^5c^5 >= a+b+c
Let
a
,
b
,
c
∈
R
+
a,b,c\in\mathbb{R^+}
a
,
b
,
c
∈
R
+
. Prove that
a
11
b
5
c
5
+
b
11
c
5
a
5
+
c
11
a
5
b
5
≥
a
+
b
+
c
\frac{a^{11}}{b^5c^5}+\frac{b^{11}}{ c^5a^5}+\frac{c^{11}}{a^5b^5}\ge a+b+c
b
5
c
5
a
11
+
c
5
a
5
b
11
+
a
5
b
5
c
11
≥
a
+
b
+
c
(Real Matrik)
7 sl9
1
Hide problems
f(x)+f(1-\frac{1}{x}) = \frac{1}{x}
Find the function
f
:
R
−
{
0
}
→
R
\displaystyle{f : \mathbb{R}-\left\{ 0\,\right\} \rightarrow \mathbb{R} }
f
:
R
−
{
0
}
→
R
such that
f
(
x
)
+
f
(
1
−
1
x
)
=
1
x
,
∀
x
∈
R
−
{
0
,
1
}
f(x)+f(1-\frac{1}{x}) = \frac{1}{x},\,\,\, \forall x \in \mathbb{R}- \{ 0, 1\,\}
f
(
x
)
+
f
(
1
−
x
1
)
=
x
1
,
∀
x
∈
R
−
{
0
,
1
}
(-InnoXenT-)
6 sl8
1
Hide problems
(1/x(s-x)^2 +1/y(s-y)^2 +1/z(s-z)^ 2)>=1/2(1/(s-x)+1/(s-y)+1/(s-z)
Let
x
,
y
,
z
x,y,z
x
,
y
,
z
represent the side lengths of any triangle, and
s
=
x
+
y
+
z
2
s=\dfrac{x+y+z}{2}
s
=
2
x
+
y
+
z
and the area of this triangle be
s
\sqrt{s}
s
square units. Prove that
s
(
1
x
(
s
−
x
)
2
+
1
y
(
s
−
y
)
2
+
1
z
(
s
−
z
)
2
)
≥
1
2
(
1
s
−
x
+
1
s
−
y
+
1
s
−
z
)
s\Big(\frac{1}{x(s-x)^2}+\frac{1}{y(s-y)^2}+\frac{1}{z(s-z)^ 2} \Big)\ge \frac{1}{2} \Big(\frac{1}{s-x}+\frac{1}{s-y}+\frac{1}{s-z}\Big)
s
(
x
(
s
−
x
)
2
1
+
y
(
s
−
y
)
2
1
+
z
(
s
−
z
)
2
1
)
≥
2
1
(
s
−
x
1
+
s
−
y
1
+
s
−
z
1
)
(Zhuge Liang)
5 sl6
1
Hide problems
x+y+z=x^2+y^2+z^2+18xyz=1
Given
x
,
y
,
z
∈
R
+
x,y,z\in \mathbb{R^+}
x
,
y
,
z
∈
R
+
. Find all sets of
x
,
y
,
z
x,y,z
x
,
y
,
z
that correspond to
x
+
y
+
z
=
x
2
+
y
2
+
z
2
+
18
x
y
z
=
1
x+y+z=x^2+y^2+z^2+18xyz=1
x
+
y
+
z
=
x
2
+
y
2
+
z
2
+
18
x
yz
=
1
(Zhuge Liang)
4 sl4
1
Hide problems
96 Thai and unknown no from Yaranakian
At the
69
69
69
Thailand-Yaranaikian meeting attended by
96
96
96
Thai delegates and a number (unknown) from the Yaranakian country. Some time after the meeting took place, the meeting also discovered something amazing that happened in this meeting!! That is, regardless of whether we select at least
69
69
69
of Thai participants and select all the Yaranikian country participants who are known to Thais in the initial selection group, there is at least
1
1
1
person fo form a minority. They found in that minority, there was always
1
1
1
more Yaranikhians than Thais. Prove that there must be at least
28
28
28
of the Yaranaikian attendees who know the Thai delegates.(Note: In this meeting, none of the attendees were half-breeds. Thai-Yara Nikian)(tatari/nightmare)
3 sl3
1
Hide problems
sum x_iy_i >= 1/n (sum x_i\)( sum y_1)
We will call the sequence of positive real numbers.
a
1
,
a
2
,
…
,
a
n
a_1,a_2,\dots ,a_n
a
1
,
a
2
,
…
,
a
n
of length
n
n
n
when
a
1
≥
a
1
+
a
2
2
≥
⋯
≥
a
1
+
a
2
+
⋯
+
a
n
n
.
a_1\geq \frac{a_1+a_2}{2}\geq \dots \geq \frac{a_1+a_2+\cdots +a_n}{n}.
a
1
≥
2
a
1
+
a
2
≥
⋯
≥
n
a
1
+
a
2
+
⋯
+
a
n
.
Let
x
1
,
x
2
,
…
,
x
n
x_1,x_2,\dots ,x_n
x
1
,
x
2
,
…
,
x
n
and
y
1
,
y
2
,
…
,
y
n
y_1,y_2,\dots ,y_n
y
1
,
y
2
,
…
,
y
n
be sequences of length
n
.
n.
n
.
Prove that
∑
i
=
1
n
x
i
y
i
≥
1
n
(
∑
i
=
1
n
x
i
)
(
∑
i
=
1
n
y
i
)
.
\sum_{i = 1}^{n}x_iy_i\geq\frac{1}{n}\left(\sum_{i = 1}^{n}x_i\right)\left(\sum_{i = 1}^{n}y_i\right).
i
=
1
∑
n
x
i
y
i
≥
n
1
(
i
=
1
∑
n
x
i
)
(
i
=
1
∑
n
y
i
)
.
(tatari/nightmare)
2 sl2
1
Hide problems
f_n=[2^n\sqrt{69}]+[2^n\sqrt{96}]
For natural
n
n
n
, define
f
n
=
[
2
n
69
]
+
[
2
n
96
]
f_n=[2^n\sqrt{69}]+[2^n\sqrt{96}]
f
n
=
[
2
n
69
]
+
[
2
n
96
]
Prove that there are infinite even integers and infinite odd integers that appear in number
f
1
,
f
2
,
…
f_1,f_2,\dots
f
1
,
f
2
,
…
.(tatari/nightmare)
1 sl1
1
Hide problems
\sum_{cyc} (a^3-b^3)^2+3\sum_{cyc}(a^2-b^2)^2+6(a-b)(b-c)(c-a)(ab+ bc+ca) >= 0.
Let
a
,
b
,
c
∈
R
a,b,c \in \mathbb{R}
a
,
b
,
c
∈
R
. Prove that
∑
c
y
c
(
a
3
−
b
3
)
2
+
3
∑
c
y
c
(
a
2
−
b
2
)
2
+
6
(
a
−
b
)
(
b
−
c
)
(
c
−
a
)
(
a
b
+
b
c
+
c
a
)
≥
0.
\sum_{cyc} (a^3-b^3)^2+3\sum_{cyc}(a^2-b^2)^2+6(a-b)(b-c)(c-a)(ab+ bc+ca) \ge 0.
cyc
∑
(
a
3
−
b
3
)
2
+
3
cyc
∑
(
a
2
−
b
2
)
2
+
6
(
a
−
b
)
(
b
−
c
)
(
c
−
a
)
(
ab
+
b
c
+
c
a
)
≥
0.
(LightLucifer)