# An Inequality for Cevians And The Ratio of Circumradius and Inradius

### Solution

By the Law of Cosines, in $\Delta XYC,$

\displaystyle \begin{align} XY^2 &= XC^2+YC^2-2\cdot XC\cdot YC\cdot\cos C\\ &\ge 2\cdot XC\cdot YC\cdot(1-\cos C)=4\cdot XC\cdot YC\cdot\sin^2\frac{C}{2}. \end{align}

It follows that $\displaystyle XY\ge 2\sqrt{XC\cdot YC}\cdot\sin\frac{C}{2}.\,$ Similarly, $\displaystyle YZ\ge 2\sqrt{YA\cdot ZA}\cdot\sin\frac{A}{2}\,$ and $\displaystyle ZX\ge 2\sqrt{XB\cdot ZB}\cdot\sin\frac{B}{2}.\,$

Multiplying the three we get

$\displaystyle \small{XY\cdot YZ\cdot ZX\ge 8\sqrt{XC\cdot XB\cdot YC\cdot YA\cdot ZA\cdot ZB}\cdot\sin\frac{A}{2}\cdot\sin\frac{B}{2}\cdot\sin\frac{C}{2}}.$

Thus,

$\displaystyle\small{\frac{XB}{XY}\cdot\frac{YC}{YZ}\cdot \frac{ZA}{ZV}\le\sqrt{\frac{XB}{XC}\cdot \frac{YC}{YA}\cdot \frac{ZA}{ZB}}\cdot \frac{\displaystyle 1}{8\cdot\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=\frac{R}{2r}}$

because, by Ceva's theorem, $g=\displaystyle \frac{XB}{XC}\cdot \frac{YC}{YA}\cdot \frac{ZA}ZB.$

### Acknowledgment

The problem by Titu Andreescu has been posted by Miguel Ochoa Sanchez at the Peru Geometrico facebook group and commented with a solution (above) by Marian Dinca. I am grateful to Leo Giugiuc for bringing this problem to my attention.

Copyright © 1996-2018 Alexander Bogomolny

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