Weizenbock’s inequality

In a triangle with sides $a$, $b$, $c$, and with area $A$, the following inequality holds:

 $a^{2}+b^{2}+c^{2}\geq 4A\sqrt{3}$

The proof goes like this: if $s=\frac{a+b+c}{2}$ is the semiperimeter of the triangle, then from Heron’s formula we have:

 $A=\sqrt{s(s-a)(s-b)(s-c)}$

But by squaring the latter and expanding the parentheses we obtain:

 $16A^{2}=2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})$

Thus, we only have to prove that:

 $(a^{2}+b^{2}+c^{2})^{2}\geq 3[2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}% +c^{4})]$

or equivalently:

 $4(a^{4}+b^{4}+c^{4})\geq 4(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})$

which is trivially equivalent to:

 $(a^{2}-b^{2})^{2}+(a^{2}-c^{2})^{2}+(b^{2}-c^{2})^{2}\geq 0$

Equality is achieved if and only if $a=b=c$ (i.e. when the triangle is equilateral) .