# An Inequality with Circumradii And Distances to the Vertices

### Solution 1

Denote $MA=x,\,$ $MB=y,\,$ $MC=z;\,$ $\angle BMC=2\alpha,\,$ $\angle AMC=2\beta,\,$ $\angle AMB=2\gamma.\,$ Clearly, $0\lt\alpha,\beta,\gamma\lt\displaystyle \frac{\pi}{2}\,$ and $\alpha +\beta +\gamma=\pi.$

From the Law of Cosines in $\Delta MBC,\,$ $BC=\sqrt{y^2-2yz\cos 2\alpha+z^2}.\,$ By the Law of Sines in $\Delta MBC,\,$ $\displaystyle R_a=\frac{\sqrt{y^2-2yz\cos 2\alpha+z^2}}{2\sin 2\alpha}.\,$ As $y^2+z^2\ge 2yz,\,$

\displaystyle \begin{align} R_a &=\frac{\sqrt{y^2-2yz\cos 2\alpha+z^2}}{2\sin 2\alpha}\\ &\ge\frac{\sqrt{2yz-2yz\cos 2\alpha}}{2\sin 2\alpha}\\ &=\frac{2\sqrt{yz}\sin\alpha}{2\sin 2\alpha}=\frac{\sqrt{yz}}{2\cos\alpha}, \end{align}

such that $\displaystyle \frac{1}{R_a}\le\frac{2\cos\alpha}{\sqrt{yz}}.\,$ We have

$\displaystyle \frac{MB\cdot MC}{R_a}=\frac{yz}{R_a}\le\frac{2yz\cos\alpha}{\sqrt{yz}}=2\sqrt{yz}\cos\alpha.$

Similarly,

$\displaystyle \frac{MC\cdot MA}{R_b}\le 2\sqrt{zx}\cos\beta\\ \displaystyle \frac{MA\cdot MB}{R_c}\le 2\sqrt{xy}\cos\gamma.$

Thus,

$\displaystyle \frac{MB\cdot MC}{R_a}+\frac{MC\cdot MA}{R_b}+\frac{MA\cdot MB}{R_c}\\ \qquad\qquad\qquad\qquad\le 2\sqrt{yz}\cos\alpha+2\sqrt{zx}\cos\beta+2\sqrt{xy}\cos\gamma.$

Suffice it to show that

$2\sqrt{yz}\cos\alpha+2\sqrt{zx}\cos\beta+2\sqrt{xy}\cos\gamma\le x+y+z.$

But, according to the famous Wolselholme's inequality,

For real $x,y,z\,$ and $\alpha+\beta+\gamma=\pi,$

$yz\cos\alpha +zx\cos\beta +xy\cos\gamma\le x^2+y^2+z^2.$

Replacing here $x,y,z\,$ with $\sqrt{x},\sqrt{y},\sqrt{z}\,$ completes the proof.

### Solution 2

Based on the following configuration and the formula $abc=4RS,$

\displaystyle\begin{align} \frac{MB\cdot MC}{R_a}&=\frac{MB\cdot MC}{\displaystyle MB\cdot MC\cdot BC\cdot\frac{1}{4[\Delta MBC]}}\\ &=4\frac{[\Delta MBC]}{BC}\\ &=2 MD, \end{align}

such that, by similarity,

\displaystyle\begin{align} &\frac{MB\cdot MC}{R_a}+\frac{MC\cdot MA}{R_b}+\frac{MA\cdot MB}{R_c}\\ &\qquad\qquad\qquad\qquad =2(MD+ME+MF)\le MA+MB+MC, \end{align}

by the Erdös-Mordell inequality.

### Acknowledgment

The problem due to Leo Giugiuc and Kadir Altintas has been posted by Leo Giugiuc at the CutTheKnotMath facebook page, with their solution (Solution 1) communicated privately. Solution 2 is by Dan Sitaru.