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.

an application of Hwalka's inequality

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 $p=vz-v.$


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