# A Cyclic Inequality in Three Variables XXII

### Solution 1

\displaystyle \begin{align} \sum_{cycl}\frac{a^3}{a^2+ab+b^2} &=\sum_{cycl}\left(a-\frac{ab(a+b)}{a^2+ab+b^2}\right)\\ &=\sum_{cycl}\left(a-\frac{a+b}{\displaystyle \frac{a}{b}+1+\frac{b}{a}}\right)\\ &\ge\sum_{cycl}\left(a-\frac{a+b}{3}\right)\\ &=(a+b+c)-\frac{2}{3}(a+b+a)\\ &=\frac{a+b+c}{3}. \end{align}

### Solution 2

Clearly, $\displaystyle \sum_{cycl}\frac{a^3}{a^2+ab+b^2}=\sum_{cycl}\frac{b^3}{a^2+ab+b^2}.\,$ Hence, we need to prove that

$\displaystyle \sum_{cycl}\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{2}{3}(a+b+c).$

But $3(a^2-ab+b^2)\ge a^2+ab+b^2,\,$ such that $\displaystyle \frac{1}{a^2+ab+b^2}\ge\frac{1}{3(a^2-ab+b^2)},\,$ from which

$\displaystyle \sum_{cycl}\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{1}{3}\sum_{cycl}\frac{a^3+b^3}{a^2-ab+b^2}=\frac{1}{3}\sum_{cycl}(a+b)=\frac{2(a+b+c)}{3}.$

### Solution 3

Since $\displaystyle \frac{a^2+b^2}{a^2+ab+b^2}-\frac{2}{3}=\frac{(a-b)^2}{3(a^2+ab+b^2)},\,$ we have

$\displaystyle \frac{a^2+b^2}{a^2+ab+b^2}\ge\frac{2}{3}.$

Using that,

\displaystyle \begin{align} \frac{a^3}{a^2+ab+b^2} &= \frac{a^3}{a^2+ab+b^2}+b -b\\ &=\frac{a^3+a^2b+ab^2+b^3}{a^2+ab+b^2}-b\\ &= \frac{(a+b)(a^2+b^2)}{a^2+ab+b^2}-b\\ &\ge\frac{2}{3}(a+b)-b=\frac{2a-b}{3}. \end{align}

It follows that

\displaystyle\begin{align}\sum_{cycl}\frac{a^3}{a^2+ab+b^2}&\ge\sum_{cycl}\frac{2a-b}{3}\\ &=\sum_{cycl}\frac{2a}{3}-\sum_{cycl}\frac{b}{3}=\sum_{cycl}\frac{a}{3}. \end{align}

### Acknowledgment

This is Problem 8.1 (high school) from the XIX (1997) Tournament of Towns. Solution 2 is by Leo Giugiuc; Solution 3 is by Marian Cucoaneş.

### References

1. L. E. Mednikov, A. V. Shapovalov, Tournaments of Towns: World of Mathematics through Problems, MCNMO, 2016 (in Russian)