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The classical counter-example to Hlawka inequality $$|a+b|+|b+c|+|c+a|\le|a+b+c|+|a|+|b|+|c|$$ is the $l^\infty$-norm in dimension $3$, with vectors $$a=\begin{pmatrix} 1 \\ 1 \\0 \end{pmatrix},\quad b=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix},\quad c=\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.$$

Since the Hlawka inequality is true in dimension one, this leaves open (?) the two-dimensional case. Actually, $\ell^\infty$ is isometric to $\ell^1$ in $2$-D, and the $\ell^1$-norm satisfies Hlawka in every space dimension.

Does every two-dimensional normed space satisfy the Hlawka inequality ?

Attempt. We may assume that $c=\alpha a+\beta b$. If the vectors $a+b$, $b+c$ and $c+a$ are convex combination of $a,b,c$ and $a+b+c$, then the convexity of the norm yields the inequality. According to my calculations, these convex combinations don't exist when $\alpha<-1$ and $\beta<0$. However, this is not a proof that Hlawka inequality fails.

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Yes, a classical result says that every 2-dimensional (real) normed space embeds into L^1.

Alternatively, if we assume that the unit ball is a $2n$-gon (the general case then follows by approximation), the corresponding space embeds into $\ell_1^n$. The dual picture is maybe even more transparent : any symmetric $2n$-gon is the Minkoswki sum of $n$ segments, and therefore is a projection of the $n$-cube.

(key word : zonoid)

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    $\begingroup$ The relevant reference is: D. Yost, $L_1$ contains every two-dimensional normed space, Ann. Polonici Math. 49 (1988), 17–19. $\endgroup$ – Tomek Kania Dec 8 '14 at 11:17

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