# A Cyclic Inequality in Three Variables XI

### Solution

First of all, since $(x-y)^2\ge 0,\,$ $x^2-xy+y^2\ge xy.\,$ It follows that $\displaystyle\frac{1}{x^2-xy+y^2}\le\frac{1}{xy}.\,$ Thus

\displaystyle\begin{align} \sum_{cycl}\frac{1}{(a^2-ab+b^2)(b^2-bc+c^2)}&\le\sum_{cycl}\frac{1}{ab}\cdot\frac{1}{bc}\\ &\le\sum_{cycl}\frac{1}{(ab)^2}\\ &\le\sum_{cycl}\frac{1}{a^2\cdot b^2}\\ &\le\sum_{cycl}\frac{1}{a^4}, \end{align}

with two applications of the Rearrangement inequality.

### Acknowledgment

Dan Sitaru has kindly posted the above problem (from the Romanian Mathematical Magazine) at the CutTheKnotMath facebook page. The Problem is credited to Nguyen Viet Hung. There you can also find a solution by Kevin Soto Palacios.