设$x,y,z$是三个不同的实数，证明
\[ \frac{1}{\max\{x,y,z\}}\leq \frac{2x\ln{x}}{(x-y)(x-z)}+\frac{2y\ln{y}}{(y-x)(y-z)}+\frac{2z\ln{z}}{(z-y)(z-x)}\leq  \frac{1}{\max\{x,y,z\}} \]
（西西）
证明：（lhclp008）
不失一般性，我们假设$x\geq y\geq z$,注意到对任意的$a,b>0$有不等式
\[ \frac{1}{\sqrt{ab}}\geq \frac{\ln{a}-\ln{b}}{a-b}\geq \frac{2}{a+b} \]
于是
\begin{align*}
\frac{1}{x}&\leq \frac{1}{x}+\frac{(\frac{2xy}{x+y}-z)(x-y)}{xy(x-z)}\\
&=\frac{\frac{4x}{x+y}-\frac{(y+z)}{y}}{(x-z)}\\
&\leq \frac{\frac{4x}{x+y}-2\frac{\sqrt{yz}}{y}}{x-z}\\
&=\frac{2x(y-z)\cdot\frac{2(x-y)}{x+y}+2z(y-x)\frac{(y-z)}{\sqrt{yz}}}{(x-y)(x-z)(y-z)}\\
&\leq \frac{2x(y-z)(\ln{x}-\ln{y})+2z(y-x)(\ln{y}-\ln{z})}{(x-y)(x-z)(y-z)}\\
&=\frac{2x\ln{x}}{(x-y)(x-z)}+\frac{2y\ln{y}}{(y-x)(y-z)}+\frac{2z\ln{z}}{(z-y)(z-x)}
\end{align*}
\begin{align*}
&\frac{2x\ln{x}}{(x-y)(x-z)}+\frac{2y\ln{y}}{(y-x)(y-z)}+\frac{2z\ln{z}}{(z-y)(z-x)}\\
&\leq \frac{2x(y-z)\frac{x-y}{\sqrt{xy}}+2z(y-x)\frac{2(y-z)}{y+z}}{(x-y)(x-z)(y-z)}\\
&=-\frac{-2z\frac{\sqrt{xy}}{y}+\frac{4z^2}{y+z}+x-z}{z(x-z)}+\frac{1}{z}\\
&\leq -\frac{-z\cdot\frac{x+y}{y}+\frac{4z^2}{y+z}+x-z}{z(x-z)}+\frac{1}{z}\\
&=-\frac{xy+xz-2yz}{yz(y+z)(x-z)}\cdot (y-z)+\frac{1}{z}\\
&\leq \frac{1}{z} 
\end{align*}

Reference:http://tieba.baidu.com/p/2388204458