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

You can use the Law of cosine and the area formula of a triangle to solve this problem. Suppose the three angles are $A,B,C$ opposite to the sides $a,b,c$, respectively. Then
$$ c^2=a^2+b^2-2ab\cos C, S=\frac{1}{2}ab\sin C $$
and hence
\begin{eqnarray*}
a^2+b^2+c^2-4S\sqrt{3}&=&2[a^2+b^2-ab(\cos C+\sqrt{3}\sin C)]\\
&=&2[a^2+b^2-2ab\sin(\frac{\pi}{6}+C)]\\
&\ge&2(a^2+b^2-2ab)\\
&=&2(a-b)^2\\
&\ge&0
\end{eqnarray*}
since $\sin(\frac{\pi}{6}+C)\le 1$. The equal sign holds iff $a=b$ and $\sin(\frac{\pi}{6}+C)=1$, which implies that $a=b, C=\frac{\pi}{3}$ or $a=b=c$.
