# An Inequality for Sides and Area

### Solution

we shall prove that for positive $x,y,$

$\displaystyle \frac{(x^2-xy+y^2)^2}{x^2+4xy+y^2}\ge\frac{1}{12}(x^2+y^2).$

To this end, introduce $s=x+y\,$ and $p=xy.\,$ Obviously, $\displaystyle \frac{s}{2}\ge \sqrt{p},\,$ implying $\displaystyle \frac{s^2}{p}\ge 4.$

We have a sequence of equivalent inequalities:

\displaystyle\begin{align} &\frac{(x^2-xy+y^2)^2}{x^2+4xy+y^2}\ge\frac{1}{12}(x^2+y^2)\\ &\frac{(s^2-3p)^2}{s^2+2p}\ge\frac{1}{12}(s^2-2p)\\ &12(s^2-3p)^2\ge s^4-4p^2\\ &12s^4-72s^2p+108p^2\ge s^4-4p^2\\ &11s^4-72s^2p+112p^2\ge 0. \end{align}

Define $\displaystyle t=\frac{s^2}{p}\,$ and note that $t\ge 4.\,$ The last inequality reduces to $11t^2-72t+112\ge 0\,$ which is the same as $(t-4)(11t-28)\ge 0,\,$ which is true since $t\ge 4.\,$ Returning to the original inequality,

\displaystyle \begin{align} \sum_{cycl}\frac{(a^2-ab+b^2)^2}{a^2+4ab+b^2}&\ge\frac{1}{12}\sum_{cycl}{a^2+b^2}\\ &=\frac{1}{6}(a^2+b^2+c^2)\ge \frac{1}{6}4\sqrt{3}S=\frac{2S}{\sqrt{3}}, \end{align}

by Weitzenböck's inequality. Equality is attained only for equilateral triangles.

### Acknowledgment

The problem (from the Romanian Mathematical Magazine) has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page, Dan later emailed me his solution in a LaTeX file.