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) .

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