# Another Refinement of the Ionescu-Weitzenböck Inequality

### Problem

### Proof

The relationship $4m_a^2=2b^2+2c^2-a^2,\;$ leads to an equivalent inequality

$\displaystyle\begin{align} a^2+b^2+c^2-4\sqrt{3}S &\ge 2\frac{2(b^2+c^2)-a^2}{4}-2h_a^2\\ &=b^2+c^2-\frac{a^2}{2}-2h_a^2, \end{align}$

which reduces to

$\displaystyle\frac{3a^2}{2}+2h_a^2\ge 4\sqrt{3}S.$

This is true because, by the AM-GM inequality,

$\displaystyle\frac{3a^2}{2}+2h_a^2 \ge 2\sqrt{\frac{3a^2}{2}\cdot 2h_a^2}=2\sqrt{3}ah_a=4\sqrt{3}S.$

The equality holds only if $\displaystyle h_a=\frac{\sqrt{3}}{2}a.$

**Note** that this does not make the triangle equilateral. Adding, say, $\displaystyle h_b=\frac{\sqrt{3}}{2}b\;$ does not yet help as the two relations hold in $\Delta ABC,\;$ with $\angle A=\angle B=30^{\circ}.\;$ However, if also $\displaystyle h_c=\frac{\sqrt{3}}{2}c,\;$ $\Delta ABC\;$ is equilateral.

### A little history

R. Weitzenböck proved the eponymous inequality in 1919. However, I. Ionescu, the founder of Romanian Mathematical Gazette, published yet in 1897 the problem:

Prove that there is no triangle for which the inequality

$4\sqrt{3}S\gt a^2+b^2+c^2$

can be satisfied.

Thus it is fair to refer to the result as the Ionescu-Weitzenböck inequality.

### Acknowledgment

The above is a small part of the article

- E. Stoica, N. Minculete, C. Barbu,
__New aspects of Ionescu–Weitzenböck’s inequality__,*Balkan Journal of Geometry and Its Applications*, Vol.21, No.2, 2016, pp. 95-101.

where among other things, the authors mention yet another refinement of Wertzenböck's inequality:

$\displaystyle a^2+b^2+c^2\ge 4S\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right)$

which is an improvement since, as is well known, $\displaystyle\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\ge\sqrt{3}.$

This appeared as *Problema 26132* by N. Minculete in *Gazeta Matematicã*, Seria B (in Romanian), 4 (2009)).

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