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.


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


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


which in the form


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.

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