An Inequality for Sides and Area

Problem

An Inequality for Sides and Area - problem

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.

 

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