An Inequality for Sides and Area


An Inequality for Sides and Area - problem


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.


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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