Weitzenböck's inequality

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Not to be confused with Weitzenböck identity. ‹See Tfd›
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.

Proofs[edit]

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[edit]

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[edit]

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} \\
\iff & & 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[edit]

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

\frac{1}{6}(a^2 + b^2 + c^2 - 4\sqrt{3}\, \Delta) ,

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

External links[edit]