A Problem From a Mongolian Olympiad for Grade 11

Problem

A Problem From a Mongolian Olympiad for Grade 11

Solution

By the AM-GM inequality,

$\begin{align} 3a+2b^3 &= (a^2+b^2+c^2)a+2b^3\\ &= a^3+ab^2+ac^2+b^3+b^3\\ &\ge 5(a^5b^8c^2)^{\frac{1}{5}}, \end{align}$

and similar for the other two fractions, so that, by the Rearrangement inequality,

$\begin{align} \sum_{cycl}\frac{a}{3a+2b^3} &\le \sum_{cycl}\frac{a}{5(a^5b^8c^2)^{\frac{1}{5}}}\\ &= \sum_{cycl}\frac{1}{5(b^8c^2)^{\frac{1}{5}}} \le \sum_{cycl}\frac{1}{5(b^8b^2)^{\frac{1}{5}}}\\ &=\sum_{cycl}\frac{1}{5(a^{10})^{\frac{1}{5}}}=\frac{1}{5}\sum_{cycl}\frac{1}{a^2}. \end{align}$

Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page the above problem from the Mongolian Mathematical Olympiad, Grade 11.

 

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