# Early Refinement of the Ionescu-Weitzenböck Inequality

### Problem

(This inequality is commonly referred to as Hadwiger-Finsler inequality. There is an additional, visually supported, proof.)

### Proof

With $s=\displaystyle(a+b+c,\,$ the semiperimeter, define $x=s-a,\,$ $y=s-b,\,$ $z=s-c.\,$ We have

\begin{align} (a^2-(b-c)^2)&+(b^2-(c-a)^2)+(c^2-(a-b)^2)\\ &=4(s-b)(s-c)+4(s-c)(s-a)+4(s-a)(s-b)\\ &=4(yz+zx+xy). \end{align}

By Heron's formula,

\begin{align} 4\sqrt{3} &= 4\sqrt{3s(s-a)(s-b)(s-c)}\\ &=4\sqrt{3(x+y+z)xyz}. \end{align}

Thus, we have to prove that $xy+yz+zx\ge\sqrt{3(x+y+z)xyz}\,.\,$ Squaring and simplifying, this becomes

$x^2y^2+y^2z^2+z^2x^2\ge x^2yz+xy^2z+xyz^2.$

This is true by the Rearrangement inequality applied to the sequence $xy,yz,zx.$

### Acknowledgment

This is a well known result, see, for example, Problems in Planimetryv II, by V. V. Prasolov (Nauka, Moscow, 1986 (Russian)).

Note that the inequality just proved is often referred to as Hadwiger-Finsler inequality.