Adil Abdulayev's Inequality With Angles, Medians, Inradius and Circumradius

Problem

Adil Abdulayev's Inequality With Angles, Medians, Inradius and Circumradius - problem

Solution

By Chebyshev's inequality,

$\displaystyle \sum_{cycl}\frac{A}{m_a}\ge\frac{1}{3}\sum_{cycl}A\cdot\sum_{cycl}\frac{1}{m_a}\frac{\pi}{3}\sum_{cycl}\frac{1}{m_a}.$

By Bergstrom's inequality,

$\displaystyle\begin{align} \frac{\pi}{3}\sum_{cycl}\frac{1}{m_a}&\ge\frac{\pi}{3}\frac{9}{\displaystyle \sum_{cycl}m_a}\ge\frac{\pi}{3}\frac{9}{4R+r}\\ &=\frac{3\pi}{4R+r}, \end{align}$

due to Leuenberger's Inequality $m_a+m_b+m_c\le 4R+r.$

Acknowledgment

Dan Sitaru has kindly posted this problem by Adil Abdulayev at the CutTheKnotMath facebook page, along with his solution. The problem was originally posted to the Romanian Mathematical Magazine.

 

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