Cyclic Inequality with Square Roots And Absolute Values

Solution

By the AM-GM inequality, $\displaystyle a+\frac{1}{3}\ge\frac{2\sqrt{a}}{\sqrt{3}},\,$ so that $\displaystyle \frac{\sqrt{3}}{2}(a+1)\ge\sqrt{a}+\frac{1}{\sqrt{3}},\,$ and, subsequently,

$\displaystyle \prod_{cycl}\frac{1}{2\sqrt{2}}(a+1)\ge\prod_{cycl}\frac{1}{\sqrt{6}}\left(\sqrt{a}+\frac{1}{\sqrt{3}}\right).$

Suffice it to show that

$\displaystyle \prod_{cycl}\left(\sqrt{a-a^2}+\frac{1}{2\sqrt{2}}|3a-1|\right)^2\ge\prod_{cycl}\frac{1}{8}(a+1)^2.$

But

\displaystyle \begin{align} \left(\sqrt{a-a^2}+\frac{1}{2\sqrt{2}}|3a-1|\right)^2&=\frac{1}{8}(a+1)^2+\frac{1}{\sqrt{2}}|3a-1|\sqrt{a-a^2}\\ &\ge\frac{1}{8}(a+1)^2, \end{align}

completing the proof.

Acknowledgment

Leo Guigiuc has kindly posted this problem at the CutTheKnotMath facebook page, with a solution of his. The problem is by Kunihiko Chikaya.

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