Weizenbock’s inequality

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

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

The proof goes like this: if $s=\frac{a+b+c}{2}$ is the semiperimeter of thetriangle, 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:

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

or equivalently:

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

which is trivially equivalent to:

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

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

See also the Hadwiger-Finsler inequality, from which this result follows as a corollary.