# An Inequality with Cotangents And the Circumradius

### Problem

### Solution

First recollect that

$\displaystyle \begin{align} &S=2R^2\sin A\sin B\sin C,\\ &\sum_{cycl}\cot A\cot B = 1,\\ &\sin^2\alpha=\frac{1}{1+\cot^2\alpha}. \end{align}$

Combined with the Law of Sines, these reduce the required inequality to $\displaystyle \sum_{cycl}\frac{xy}{(z+x)(z+y)}\le 1,\,$ where $x=\cot A,\,$ $y=\cot B,\,$ $z=\cot C.\,$ The latter inequality can be rewritten as

$\displaystyle \sum_{cycl}xy(x+y)\le 2xyz+\sum_{cycl}xy(x+y),$

which is obviously true, since $0\le xyz.\,$ Interestingly, *equality holds for all right triangles* but *not for equilateral ones*.

An example shows that the inequality is reversed for obtuse triangles: take, e.g., $a=\sqrt{3}\,$ and $b=c=1.$

### Acknowledgment

I am grateful to Leo Giugiuc for posting a problem of his from the Gazeta Matematica at the CutTheKnotMath facebook page and later commenting with his solution.

[an error occurred while processing this directive]

|Contact| |Up| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny