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

[latex]\displaystyle\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+2b+c)^2}\le\frac{3}{16}.[/latex]

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

[latex]\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}).[/latex]

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

[latex]a(1-b),b(1-c),c(1-a)[/latex]

中必有一個數不小於[latex]\frac{1}{4}[/latex].

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

[latex]\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.[/latex]

求證:[latex]a_{1}+a_{2}+\ldots+a_{100}\le 9900[/latex].

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

[latex]x_1^2+x_2^2+\cdots+x_n^2\ge (x_1+x_2+\cdots+x_{n-1})x_n[/latex]

對所有實數[latex]x_1,x_2,\cdots,x_n[/latex]成立.

問題6.Hungary-Israel Binational 2009
設[latex]x,y,z[/latex]為非負數,求證:

[latex]\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}}.[/latex]

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

[latex]x^4+y^4+z^4+16(x^2+y^2+z^2)\ge 8(x^3+y^3+z^3)+27,[/latex]

并求出等號于何時成立.

問題8.International Zhautykov Olympiad 2009
設凸六邊形[latex]ABCDEF[/latex]的面積為[latex]S[/latex],求證

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

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

[latex]\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}}[/latex]

的所有可能值(用[latex]n[/latex]表示).

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

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

[latex]\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}|.[/latex]

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

[latex]\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}[/latex]

其中[latex]a_i=a_{i+4}[/latex]對所有整數[latex]i[/latex]成立.

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

[latex]0=a_0\le a_1\le a_2\le \cdots \le a_n,[/latex]

及

[latex]\displaystyle a_i\ge\frac{a_{i+1}+a_{i-1}}{2},i=1,2,\cdots,n-1,[/latex]

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

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

問題15.China Girls Math Olympiad 2009
設[latex]x,y,z\ge 1[/latex],求證:

[latex]\displaystyle \prod (x^2-2x+2)\le x^2y^2z^2-2xyz+2.[/latex]

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

[latex]2a_k\le a_{k-1}+a_{k+1},k=2,3,\cdots,n-1.[/latex]

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

[latex]|a_k|\le \lambda(n)\max\{|a_1|,|a_n|\}.[/latex]

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

[latex]a^2+b^2+c^2\ge 2(ab+bc+ca).[/latex]

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

[latex]\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},[/latex]

試求出[latex]f(x,y,z)[/latex]的最大值和最小值.

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

[latex]\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}.[/latex]

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

問題21.Costa Rica Final Round 2009
設多項式[latex]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)[/latex],其中[latex]r_1,r_2,\cdots,r_n[/latex]為實數.求證:[latex](n-1)a_{n-1}^2\ge 2na_{n-2}.[/latex]

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

[latex]\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge 0.[/latex]

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

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

[latex]\displaystyle a+b+c+d <\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}.[/latex]

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

[latex]\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,[/latex]

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[latex]ABC[/latex]的垂心為[latex]H[/latex],令[latex]h_{\max}[/latex]表示三角形[latex]ABC[/latex]的最大高的長度.求證:

[latex]AH+BH+CH\le 2h_{\max}.[/latex]

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

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

<未完待續[latex]\cdots[/latex]>