# An Application of Hlawka's Inequality

Here's an inequality that Leo Giugiuc found at the AOPS forum and posted with a solution at the CutTheKnotMath facebook page. The elegant solution is by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc.

For $a,\;b,\;c\gt 0$ the following inequality holds:

\begin{align} \sqrt{a^2-ab+b^2}&+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\\ &\le a+b+c+\sqrt{a^2+b^2+c^2-ab-bc-ca}. \end{align}

### Solution

Set $\displaystyle w=-\frac{1}{2}+i\frac{\sqrt{3}}{2},$ which is a rotation by $120^{\circ}$ counterclockwise. Define $x=a,$ $y=bw,$ $z=cw^2.$

First of all $|x|=|a|,$ $|y|=|b|,$ $|z|=|c|.$ Then also, $|x+y|=\sqrt{a^2-ab+b^2},$ $|y+z|=\sqrt{b^2-bc+c^2},$ $|z+x|=\sqrt{c^2-ca+a^2},$ and $|x+y+z|=\sqrt{a^2+b^2+c^2-ab-bc-ca}.$ I'll verify the latter identity:

$\displaystyle x+y+z=\left(a-\frac{1}{2}b-\frac{1}{2}c\right)+i\left(\frac{\sqrt{3}}{2}b-\frac{\sqrt{3}}{2}c\right).$

It follows that

\displaystyle\begin{align} |x+y+z|^2&=\left(a-\frac{1}{2}b-\frac{1}{2}c\right)^2+\frac{3}{4}(b-c)^{2}\\ &=a^2+\frac{1}{4}b^2+\frac{1}{4}c^2-ab-ac+\frac{1}{2}bc+\frac{3}{4}b^2+\frac{3}{4}c^2-\frac{3}{2}bc\\ &=a^2+b^2+c^2-ab-bc-ca, \end{align}

as required. It is now clearly seen that the problem is simply a reformulation of Hlawka's inequality:

$|x+y|+|y+z|+|z+x|\le |x|+|y|+|z|+|x+y+z|,$

true for any three complex numbers $x,\;y,\;z.$