# Inequality in Acute Triangle

Leo Giugiuc posted an elegant problem and its solution at the CutTheKnotMath facebook page. The problem and the solution are due to Dan Sitaru and Leo Giugiuc.

In the usual notations,

$a+b+c\ge\sqrt{2bc\cos A}+\sqrt{2ac\cos B}+\sqrt{2ab\cos C}.$

with equality only in equilateral triangle.

### Proof

By the Law of Cosines, $2bc\cos A=b^2+c^2-a^2,$ so that the required inequality can be rewritten as

$a+b+c\ge\sqrt{b^2+c^2-a^2}+\sqrt{a^2+c^2-b^2}+\sqrt{a^2+b^2-c^2}.$

Let's make a substitution: $a=\sqrt{y+z},$ $b=\sqrt{x+z},$ $c=\sqrt{x+y}$ which converts the inequality to

$\sqrt{y+z}+\sqrt{x+z}+\sqrt{x+y}\ge\sqrt{2x}+\sqrt{2y}+\sqrt{2z},$

which in the form

$\displaystyle\sqrt{\frac{y+z}{2}}+\sqrt{\frac{x+z}{2}}+\sqrt{\frac{x+y}{2}}\ge\sqrt{x}+\sqrt{y}+\sqrt{z},$

follows from a particular case of Jensen's inequality, $\displaystyle\sqrt{\frac{s+t}{2}}\ge\frac{\sqrt{s}+\sqrt{t}}{2},$ or by the Arithmetic Mean-Quadratic Mean inequality.

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny