# Ji Chen's Inequality

### Solution

Write $a=y+z,$ $b=z+x,$ $c=x+y.$ Then $\displaystyle x=\frac{b+c-a}{2},$ etc., and

\displaystyle\begin{align}\ \sum_{cycl}yz&=\sum_{cycl}\frac{a^2-(b-c)^2}{4}\\ &=\sum_{cycl}\frac{a^2-b^2-c^2+2bc}{4}=\frac{1}{4}\sum_{cycl}(2bc-a^2). \end{align}

WLOG, assume $a\ge b\ge c;$ then

$2bc-a^2\le 2ca-b^2\le 2ab-c^2.$

Therefore, by Chebyshev's inequality and then by the AM-GM inequality,

\displaystyle\begin{align} \frac{4}{9}\left(\sum_{cycl}yz\right)\left(\sum_{cycl}\frac{1}{(y+z)^2}\right)&=\frac{1}{3}\sum_{cycl}(2bc-a^2)\cdot\frac{1}{3}\sum_{cycl}\frac{1}{a^2}\\ &\ge\frac{1}{3}\sum_{cycl}\left((2bc-a^2)\cdot\frac{1}{a^2}\right)\\ &=\frac{1}{3}\left(\sum_{cycl}\frac{2bc}{a^2}\right)-1\\ &\ge\left(\prod_{cycl}\frac{2bc}{a^2}\right)^{\frac{1}{3}}-1=2-1=1 \end{align}

and the required inequality follows.

### Acknowledgment

This is problem 1940 from the Canadian Crux Mathematicorum (April 1994). Solution is by Marcin E. Kuczma (March 1995).