問題1.IMO Shortlist 2009
設正數$a,b,c$滿足$\displaystyle a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$,求證:

$\displaystyle\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+2b+c)^2}\le\frac{3}{16}.$

問題2.IMO Shortlist 2009
已知$a,b,c>0$且$ab+bc+ca\leq 3abc$.求證:

$\displaystyle \sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+\sqrt{\frac{a^2+b^2}{a+b}}+3\le \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}).$

問題3.Junior Balkan MO 2009
設實數$a,b,c$滿足$0<a,b,c<1$且$abc=(1-a)(1-b)(1-c)$,求證:

$a(1-b),b(1-c),c(1-a)$

中必有一個數不小於$\frac{1}{4}$.

問題4.Baltic Way 2009
設非負整數$a_{1},a_{2},\cdots ,a_{100}$滿足

$\displaystyle\sum_{i=1}^{100}a_{i}\cdot (a_{i}-1)\cdot\ldots\cdot (a_{i}-20)\le 100\cdot 99\cdot 98\cdot\ldots\cdot 79.$

求證:$a_{1}+a_{2}+\ldots+a_{100}\le 9900$.

問題5.Baltic Way 2009
求出所有大於的 $1$ 的整數$n$使得不等式

$x_1^2+x_2^2+\cdots+x_n^2\ge (x_1+x_2+\cdots+x_{n-1})x_n$

對所有實數$x_1,x_2,\cdots,x_n$成立.

問題6.Hungary-Israel Binational 2009
設$x,y,z$為非負數,求證:

$\displaystyle\frac{x^{2}+y^{2}+z^{2}+xy+yz+zx}{6}\le\frac{x+y+z}{3}\cdot\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}.$

問題7.Middle European Mathematical Olympiad 2009
設$x,y,z$為滿足$x^2+y^2+z^2+9=4(x+y+z)$的實數,求證:

$x^4+y^4+z^4+16(x^2+y^2+z^2)\ge 8(x^3+y^3+z^3)+27,$

并求出等號于何時成立.

問題8.International Zhautykov Olympiad 2009
設凸六邊形$ABCDEF$的面積為$S$,求證

$AC\cdot(BD+BF-DF)+CE\cdot(BD+DF-BF)+AE\cdot(BF+DF-BD)\geq 2\sqrt{3}S.$

問題9.Brazil National Olympiad 2009
給定整數$n>3$,$x_1,x_2,\cdots,x_n$為正實數,試求出表達式

$\displaystyle {x_{1}\over x_{n}+x_{1}+x_{2}}+{x_{2}\over x_{1}+x_{2}+x_{3}}+{x_{3}\over x_{2}+x_{3}+x_{4}}+\cdots+{x_{n-1}\over x_{n-2}+x_{n-1}+x_{n}}+{x_{n}\over x_{n-1}+x_{n}+x_{1}}$

的所有可能值(用$n$表示).

問題10.China National Olympiad 2009
給定正整數$n\ge 3$,實數$a_1,a_2,\cdots,a_n$滿足$\displaystyle \min_{1\le i<j\le n}\left|a_i-a_j\right|=1$.求$\displaystyle\sum_{k=1}^n \left|a_k\right|^3$的最小值.

問題11.China Team Selection Test 2009
已知$x_1,x_2,\cdots,x_m,y_1,y_2,\cdots,y_n$是正實數,記$\displaystyle X=\sum_{i=1}^n x_i,Y=\sum_{i=1}^n y_i.$求證:

$\displaystyle 2XY\sum_{i = 1}^{m}\sum_{j = 1}^{n}|x_{i}-y_{j}|\ge X^{2}\sum_{j = 1}^{n}\sum_{l = 1}^{n}|y_{i}-y_{l}|+Y^{2}\sum_{i = 1}^{m}\sum_{k = 1}^{m}|x_{i}-x_{k}|.$

問題12.China Team Selection Test 2009
非負實數$a_1,a_2,a_3,a_4$滿足$a_1+a_2+a_3+a_4=1$.求證:

$\begin{aligned}\max\Big\{& \sum_{i=1}^{4}{\sqrt{a_{i}^{2}+a_{i}a_{i-1}+a_{i-1}^{2}+a_{i-1}a_{i-2}}},\\ & \sum_{i=1}^{4}{\sqrt{a_{i}^{2}+a_{i}a_{i+1}+a_{i+1}^{2}+a_{i+1}a_{i+2}}}\Big\}\ge 2,\end{aligned}$

其中$a_i=a_{i+4}$對所有整數$i$成立.

問題13.China Team Selection Test 2009
給定整數$n\ge 2$,求具有下列性質的最大常數$\lambda(n)$:若實數序列$a_0,a_1,a_2,\cdots,a_n$滿足

$0=a_0\le a_1\le a_2\le \cdots \le a_n,$

及

$\displaystyle a_i\ge\frac{a_{i+1}+a_{i-1}}{2},i=1,2,\cdots,n-1,$

則有$\displaystyle\left(\sum_{i=1}^n ia_i\right)^2\ge \lambda(n)\sum_{i=1}^n a_i^2$.

問題13.China Team Selection Test 2009
設整數$m\ge 2$,$n$為奇數且$3\le n<2m$.數$a_{i,j}(i,j\in \mathbb{N},1\le i\le m,1\le j\le n)$滿足:
(1)對於任意的$1\le j\le n,a_{1,j},a_{2,j},\cdots,a_{m,j}$是$1,2,\cdots,m$的一個排列.
(2)對於任意的$1\le i\le m,1\le j\le n-1,$有$\left|a_{i,j}-a_{i,j+1}\right|\le 1$.
求$\displaystyle M=\max_{1\le i\le m}\sum_{j=1}^n a_{i,j}$的最小值.

問題15.China Girls Math Olympiad 2009
設$x,y,z\ge 1$,求證:

$\displaystyle \prod (x^2-2x+2)\le x^2y^2z^2-2xyz+2.$

問題16.China Western Mathematical Olympiad 2009
設實數$a_{1},a_{2},\cdots ,a_{n}(n\ge 3)$滿足$a_{1}+a_{2}+\cdots+a_{n}=0$,且

$2a_k\le a_{k-1}+a_{k+1},k=2,3,\cdots,n-1.$

求最小的$\lambda(n)$,使得對所有的$k\in \{1,2,\cdots,n\}$,都有

$|a_k|\le \lambda(n)\max\{|a_1|,|a_n|\}.$

問題17.China South East Mathematical Olympiad 2009
設$x,y,z$為正實數,令$\sqrt{a}=x(y-z)^2,\sqrt{b}=y(z-x)^2,\sqrt{c}=z(x-y)^2$.求證:

$a^2+b^2+c^2\ge 2(ab+bc+ca).$

問題18.China South East Mathematical Olympiad 2009
設非負實數$x,y,z$滿足$x+y+z=1$,設設$f(x,y,z)$定義如下:

$\displaystyle�f(x,y,z)=\frac{x(2y-z)}{1+x+3y}+\frac{y(2z-x)}{1+y+3z}+\frac{z(2x-y)}{1+z+3x},$

試求出$f(x,y,z)$的最大值和最小值.

問題19.China Northern Mathematical Olympiad
設正實數$x,y,z$滿足$x^2+y^2+z^2=3$,求證:

$\displaystyle\frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2009}-2008(y-1)}{z+x}+\frac{z^{2009}-2008(z-1)}{x+y}\ge\frac{x+y+z}{2}.$

問題20.Costa Rica Final Round 2009
設正實數$x,y$滿足$(1+x)(1+y)=2$,求證:$\displaystyle�xy+\frac{1}{xy}\ge 6.$

問題21.Costa Rica Final Round 2009
設多項式$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=(x+r_1)(x+r_2)\cdots(x+r_n)$,其中$r_1,r_2,\cdots,r_n$為實數.求證:$(n-1)a_{n-1}^2\ge 2na_{n-2}.$

問題22.Croatia MEMO Team Selection Tests 2009
求證對任意的正實數$a,b,c,d$都有

$\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge 0.$

問題23.Croatia MEMO Team Selection Tests 2009
三角形$ABC$的邊$AB,AC$上分別有一點$D,E$,滿足$DE$是三角形$ABC$內切圓的切線,且$DE$平行於$BC$.求證:$AB+BC+CA\ge 8DE$.

問題24.Germany Team Selection Tests 2009
設正實數$a,b,c,d$滿足正實數$abcd=1,a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$.求證:

$\displaystyle�a+b+c+d <\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}.$

問題25.Germany Team Selection Tests 2009
證明對任意正實數$a,b,c,d$都有

$\displaystyle\frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b}\ge 0,$

并求所有等號成立的可能情形.

問題26.India National Olympiad 2009
設銳角三角形$ABC$的垂心為$H$,令$h_{\max}$表示三角形$ABC$的最大高的長度.求證:

$AH+BH+CH\le 2h_{\max}.$

問題27.India National Olympiad 2009
正實數$a,b,c$滿足$a^3+b^3=c^3$,求證:$a^{2}+b^{2}-c^{2}> 6(c-a)(c-b)$.

問題28.India International Mathematical Olympiad Training Camp 2009
設$P$三角形$ABC$內部一點,求證:$\displaystyle\frac{PA}{a}+\frac{PB}{b}+\frac{PC}{c}\ge\sqrt{3}$.

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