Geometric Proof of Hlawka's Inequality

Problem

Geometric Proof of Hlawka's Inequality, problem

Solution

Laying off successively vectors $\vec{MN}=\vec{a},\,$ $\vec{NP}=\vec{b},\,$ $\vec{PQ}=\vec{c},\,$ $\vec{QM}=\vec{d}\,$ we obtain a closed broken line - a polygon, because $\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}.\,$ Ordering the vectors differently will result in a different polygon. We are going to prove that among those polygons there are self-intersecting.

Choose of of the polygons randomly. If one of its vertices, say $Q,\,$ lies within $\Delta MNP,\,$ add vector $\vec{QQ'}=\vec{MN}.\,$ Since, obviously, $\vec{Q'N}=\vec{QM},\,$ the polygon $NPQQ'\,$ is a legitimate self-intersecting member of our collection of polygons.

Geometric Proof of Hlawka's Inequality, #1

In case $MNPQ\,$ is a convex polygon, we first do the construction illustrated below:

Geometric Proof of Hlawka's Inequality, #2

Thus assume polygon $MNPQ\,$ is self-intersecting and, more specifically, let $R\,$ be the intersection of $NP\,$ and $QM.$

Geometric Proof of Hlawka's Inequality, #3

Then, $|MR|+|PR|\ge |PM|,\,$ $|NR|+|QR|\ge |QN|,\,$ implying

$|MR|+|PR|+|NR|+|QR|\ge |PM|+|QN|.$

But

$|PM|=|\vec{PM}|=|\vec{PQ}+\vec{QM}|=|\vec{c}+\vec{d}|,\\ |QN|=|\vec{QN}|=|\vec{QM}+\vec{MN}|=|\vec{a}+\vec{d}|,$

while $|MR|+|PR|+|NR|+|QR|=|QM|+|NP|=|\vec{b}+\vec{d}|.\,$ It follows that

(1)

$|\vec{b}|+|\vec{d}|\ge |\vec{a}+\vec{d}|+|\vec{c}+\vec{d}|.$

In addition, $|\vec{a}|+|\vec{c}|\ge |\vec{a}+\vec{c}|\,$ and $|\vec{a}+\vec{c}|=|\vec{b}+\vec{d}|.\,$ Hence,

(2)

$|\vec{a}|+|\vec{c}|\ge |\vec{b}+\vec{d}|.$

Adding up (1) and (2), we obtain the required inequality.

Acknowledgment

The above problem which is actually a reformulation of Hlawka's inequality has been published in the popular Russian magazine Kvant as Problem M394 with the solution by Yu. Ionin (Kvant, 1977, 03, p 32)

Since $\vec{d}=-\vec{a}-\vec{b}-\vec{c}\,$ and, e.g., $\vec{d}+\vec{a}=-\vec{b}-\vec{c},\,$ the above inequality is equivalent to Hlawka's:

$|\vec{a}|+|\vec{b}|+|\vec{c}|+|\vec{a}+\vec{b}+\vec{c}|\ge |\vec{b}+\vec{c}|+|\vec{c}+\vec{a}|+|\vec{a}+\vec{b}|.$

 

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