# Seyran Ibrahimov's Inequality

### Solution

We'll employ Leuenberger's inequality $\displaystyle \sum_{cycl}m_a\le 4R+r\,$ and Mitrinović's inequality $\displaystyle \sum_{cycl}a\le 3\sqrt{3}R:$

\displaystyle \begin{align} \sqrt{3}s\cdot\sum_{cycl}m_a &\le \sqrt{3}\cdot (4R+r)\cdot \frac{1}{2}3\sqrt{3}R\\ &=\frac{36R^2+9Rr}{2}. \end{align}

Suffice it to prove that $\displaystyle \frac{36R^2+9Rr}{2}\le 20R^2+r^2.\,$ The latter is equivalent to $4R^2-9Rr=2r^2\ge 0,\,$ i.e., $(R-2r)(4R-r)\ge 0\,$ which is true due to Euler's inequality $R\ge 2r.$.

### Acknowledgment

Dan Sitaru has kindly posted the problem and his solution at the CutTheKnotMath facebook page and helped me out when I lost the link. I am deeply in Dan's debt. The problem is by Seyran Ibrahimov. George Apostolopoulos gave the same solution. This problem of his was originally published at the Romanian Mathematical Magazine.

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