Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that
$$4S \sqrt{3}\le a^2+b^2+c^2$$ 

Let $\triangle$ denote the area.
$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$
Take $b+c-a=x$
$c+a-b=y$ and $a+b-c=z$.
$\dfrac{x}{2}=(s-a)$
$\dfrac{y}{2}=(s-b)$
$\dfrac{z}{2}=(s-c)$
$\dfrac{x+y+z}{2}=s$
Now use AM-GM inequality.
$\dfrac{x+y+z}{4} \ge ((xyz)(x+y+z))^\frac{1}{4}$
$\dfrac{2}{x+y+z} \le \dfrac{1}{((xyz)(x+y+z))^\frac{1}{4}}$
$\dfrac{4}{(x+y+z)^2} \le \dfrac{1}{((xyz)(x+y+z))^\frac{1}{2}} \implies \dfrac{4}{(x+y+z)^2} \le \dfrac{1}{\triangle}$
