# Does Hlawka Inequality follow from Triangle Inequality?

On MathOverflow I saw this inequality. Let $E$ is a normed linear space.

$$\|x+y\|+\|y+z\|+\|z+x\|\le\|x\|+\|y\|+\|z\|+\|x+y+z\|,\qquad\forall x,y,z\in E$$

Apparently this is always true if $E = \mathbb{R}^n$. I always thought Triangle inequality was the only foundation inequality for normed spaces. Does Hlawka inequality follow as a consequence?

The same MO question discusses potential counterexamples. Why is the space of $2 \times 2$ self-adjoint matrices not Euclidean with the following norm?

$$||A|| = \tfrac{1}{2}|\mathrm{tr}( A)| + \tfrac{1}{\sqrt{2}}\big|\big| A - \mathrm{tr}(A)I_{2\times 2} \big|\big|_2$$

The weighted sum of the trace and the Hilbert-Schmidt norm.

• For references, see also here. – Dietrich Burde Nov 3 '14 at 21:20
• The Hlawka Identity takes care of it, $\displaystyle (\|x\|+\|y\|+\|z\|-\|x+y\|-\|y+z\|-\|z+x\|+\|x+y+z\|)(\|x\|+\|y\|+\|z\| + \|x+y+z\|) = \sum\limits_{cyc} (\|x\|+\|y\| - \|x+y\|)(\|z\| - \|x+y\| + \|x+y+z\|)$ – r9m Nov 3 '14 at 21:28

Counterexample:in $\mathbb{R}^3$, with norm defined by $(x,y,z)_p=(|x|^p+|y|^p+|z|^p)^{1/p}$, $p\ge 3$. Take $x=(1,1,-1)$, $y=(-1,1,1)$, $z=(1,-1,1)$. Clearly the inequality holds for $p=1$ and $p=2$. It was proved that any finite two dimesional space satisfies this inequality (and so is Hlawka space).