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
$$4S \sqrt{3}\le a^2+b^2+c^2$$ 

$\dfrac{S^{2}}{s}=(s-a)(s-b)(s-c)\leq \left(\dfrac{(s-a)+(s-b)+(s-c)}{3}\right)^{3}=\dfrac{s^{3}}{27}$
$\therefore$ $S\leq\dfrac{s^{2}}{3\sqrt{3}}=\dfrac{(a+b+c)^{2}}{12\sqrt{3}}\leq\dfrac{1}{12\sqrt{3}}\cdot 3(a^{2}+b^{2}+c^{2})$
