# 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.