# Weitzenböck's inequality

Not to be confused with Weitzenböck identity.
According to Weitzenböck's inequality, the area of this triangle is at most (a2 + b2 + c2) ⁄ 4√3.

In mathematics, Weitzenböck's inequality, named after Roland Weitzenböck, states that for a triangle of side lengths $a$, $b$, $c$, and area $\Delta$, the following inequality holds:

$a^2 + b^2 + c^2 \geq 4\sqrt{3}\, \Delta.$

Equality occurs if and only if the triangle is equilateral. Pedoe's inequality is a generalization of Weitzenböck's inequality. The Hadwiger-Finsler inequality is a strengthened version of Weitzenböck's inequality.

## Proofs

The proof of this inequality was set as a question in the International Mathematical Olympiad of 1961. Even so, the result is not too difficult to derive using Heron's formula for the area of a triangle:

\begin{align} \Delta & {} = \frac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4} \\ & {} = \frac{1}{4} \sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}. \end{align}

### First method

This method assumes no knowledge of inequalities except that all squares are nonnegative.

\begin{align} {} & (a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2 \geq 0 \\ {} \iff & 2(a^4+b^4+c^4) - 2(a^2 b^2+a^2c^2+b^2c^2) \geq 0 \\ {} \iff & \frac{4(a^4+b^4+c^4)}{3} \geq \frac{4(a^2 b^2+a^2c^2+b^2c^2)}{3} \\ {} \iff & \frac{(a^4+b^4+c^4) + 2(a^2 b^2+a^2c^2+b^2c^2)}{3} \geq 2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4) \\ {} \iff & \frac{(a^2 + b^2 + c^2)^2}{3} \geq (4\Delta)^2, \end{align}

and the result follows immediately by taking the positive square root of both sides. From the first inequality we can also see that equality occurs only when $a = b = c$ and the triangle is equilateral.

### Second method

This proof assumes knowledge of the rearrangement inequality and the arithmetic-geometric mean inequality.

\begin{align} & & a^2 + b^2 + c^2 & \geq & & ab+bc+ca \\ \iff & & 3(a^2 + b^2 + c^2) & \geq & & (a + b + c)^2 \\ \iff & & a^2 + b^2 + c^2 & \geq & & \sqrt{3 (a+b+c)\left(\frac{a+b+c}{3}\right)^3} \\ \Rightarrow & & a^2 + b^2 + c^2 & \geq & & \sqrt{3 (a+b+c)(-a+b+c)(a-b+c)(a+b-c)} \\ \iff & & a^2 + b^2 + c^2 & \geq & & 4 \sqrt3 \Delta. \end{align}

As we have used the rearrangement inequality and the arithmetic-geometric mean inequality, equality only occurs when $a = b = c$ and the triangle is equilateral.

### Third method

It can be shown that the area of the inner Napoleon's triangle, which must be nonnegative, is[1]

$\frac{\sqrt{3}}{24}(a^2+b^2+c^2-4\sqrt{3}\Delta),$

so the expression in parentheses must be greater than or equal to 0.