IMO 1961 - Problem 2

Problem: Let $a,b$ and $c$ be the lengths of a triangle whose area is $S$. Prove that $a^2+b^2+c^2\ge 4S\sqrt{3}$
Solution: Using $S=\frac{bcsin(\alpha)}{2},a^2=b^2+c^2-2bccos(\alpha)$ and $\frac{\sqrt{3}sin(\alpha)+cos(\alpha)}{2}=cos(\alpha-60^0)$ we have
$a^2+b^2+c^2\ge 4S\sqrt{3}\iff b^2+c^2\ge bc(\sqrt{3}sin(\alpha)+cos(\alpha))\iff (b-c)^2+2bc(1-cos(\alpha-60^0))\ge 0$,
where equality holds if and only if $b=c$ and $\alpha=60^0$, i.e, if the triangle is equilateral.