An Inequality with Powers of Six

Problem

An Inequality with Powers of Six

Proof

From $(a-b)^2\ge 0,\,$ $a^2-ab+b^2\ge ab.\,$ So, using Euler's inequality $R\ge 2r,\,$ $abc=4RS\,$ and $S=rs$,

$\displaystyle\begin{align} RHS &= 8r^2s\sum_{cycl}\frac{a^5}{b^2-bc+c^2}\le 8r^2s\sum_{cycl}\frac{a^5}{bc}\\ &\le\frac{8r^2s}{abc}\left(\sum_{cycl}a^6\right)=\frac{8r^2s}{4RS}\left(\sum_{cycl}a^6\right)\\ &=\sum_{cycl}a^6 = LHS. \end{align}$

Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page the above problem from his book Math Accent, with a proof Diego Alvariz (India); Seyran Ibrahimov (Azerbaijan) submitted the same proof independently.

 

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