Weizenbock's inequality

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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.

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Theorem

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