# Hlawka's Inequality

The subject of the page is to establish an inequality named after Edmund Hlawka (1916-2009), an Austrian mathematician.

For complex $x,y,z,$

(*)

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

The inequality holds in vector spaces with scalar product. Both proofs below may be extended to the general case.

### Proof 1

The inequality (*) is equivalent to

$|x|+|y|+|z|-|x+y|-|y+z|-|z+x|+|x+y+z|\ge 0.$

Multiply that by $|x|+|y|+|z| + |x+y+z|$ and check that after straightforward algebraic manipulation the product reduces to

\begin{align}\displaystyle &(|x|+|y| - |x+y|)(|z| - |x+y| + |x+y+z|)+\\ &(|y|+|z| - |y+z|)(|x| - |y+z| + |x+y+z|)+\\ &(|z|+|x| - |z+x|)(|y| - |z+x| + |x+y+z|)\ge 0. \end{align}

By the triangle inequality in each of the three terms, both factors are not negative, and this proves the inequality. This also tells us that the equality is only reached when each of the three terms vanishes. For this to happen $x,y,z$ need to be collinear with the origin.

### Proof 2

Both sides of the inequality (*) are positive so that we may replace it with the squared one. For the left-hand side we have

\begin{align} (|x+y|+|y+z|+|z+x|)^{2} &= 2(x^{2}+y^{2}+z^{2})+2(|xy|+|yz|+|zx|)\\ &+2(|x+y|\cdot |y+z|+|y+z|\cdot |z+x|+|z+x|\cdot |x+y|). \end{align}

For the right side we obtain

\begin{align} (|x|+|y|+|z|+|x+y+z|)^{2} &= 2(x^{2}+y^{2}+z^{2})+2(|xy|+|yz|+|zx|)\\ &+2(|x|+|y|+|z|)|x+y+z|+2(|xy|+|yz|+|zx|). \end{align}

Thus (*) can be rewritten as

\begin{align} |x+y|&\cdot |y+z|+|y+z|\cdot |z+x|+|z+x|\cdot |x+y|\\ &\le (|x|+|y|+|z|)|x+y+z|+(|xy|+|yz|+|zx|). \end{align}

For the proof, regroup the terms:

\begin{align} &(|xy|+|z(x+y+z)|-|y+z|\cdot |z+x|)+\\ &(|yz|+|x(x+y+z)|-|z+x|\cdot |x+y|)+\\ &(|zx|+|y(x+y+z)|-|x+y|\cdot |y+z|)\ge 0. \end{align}

Note that each of the terms in the latter sum is not negative because, for example,

$|xy|+|z(x+y+z)|\ge |xy+z(x+y+z)|=|(x+z)(y+z)|.$

The equality is achieved only when $xy$ and $z(x+y+z)$ are collinear with the origin and are on the same ray from the origin, or, when

$\displaystyle\frac{z(x+y+z)}{xy}\gt 0.$

More inclusively,

$\overline{xy}z(x+y+z)\ge 0.$

For (*) to become equality, the three inequalities (the latter one and the two obtained by permutation of the variables) need to be equalities.

### Acknowledgment

Proof 2 has been supplied by Leo Giugiuc; Proof 1 I have lifted from a discussion at MathOverflow. A third (geomteric) proof can be found on a separate page.

Hlawka's Inequality has a very nice application.

Copyright © 1996-2018 Alexander Bogomolny

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