# An Application of Hlawka's Inequality II

Here's an inequality due to Nguyen Hung Viet that was posted by Leo Giugiuc at the CutTheKnotMath facebook page along with the solution.

### Solution

WLOG, let $A=(0,\sqrt{3}),\;$ $B=(-1,0),\;$ $C=(1,0),\;$ meaning, in particular, that $a=2.\;$ Let $M=(-u+v, w\sqrt{3}),\;$ with $u,v,w\gt 0,\;$ and $u+v+w=1.$

From these we easily deduce that $x=w\sqrt{3},\;$ $y=u\sqrt{3}\;$ and $z=v\sqrt{3}.$

We'll work with complex numbers so that $A=i\sqrt{3},\;$ $B=-1,\;$ $C=1,\;$ $P=z,\;$ $M=-u+v+wi\sqrt{3};\;$ $\displaystyle A'=\frac{-u+v}{u+v},\;$ $\displaystyle B'=\frac{v}{v+w}+\frac{w\sqrt{3}}{v+w},\;$ $\displaystyle C'=-\frac{u}{u+w}+\frac{w\sqrt{3}}{u+w}i.\;$

Hlawka's inequality tells us that $|m|+|n|+|p|+|m+n+p|\ge |m+n|+|m+p|+|n+p|.\;$ We need to prove that

\begin{align}w|z-i\sqrt{3}|&+u|z+1|+v|z-1|+|z+u-v-wi\sqrt{3}|\ge\\ &|(u+v)z+u-v|+|(v+w)z-v-wi\sqrt{3}|+|(u+w)z+u-wi\sqrt{3}|. \end{align}

To obtain that, we apply Hlawka's inequality above with $m=wz-wi\sqrt{3},\;$ $n=uz+u\;$ and $p=vz-v.$