Operator norm versus Hlawka inequality Let $E$ be a finite dimensional normed vector space. If $E$ is $\ell^1$-embeddable, then the norm satisfies Hlawka inequality
$${\bf(H)}\qquad\|x+y\|+\|y+z\|+\|z+x\|\le\|x\|+\|y\|+\|z\|+\|x+y+z\|,\qquad\forall x,y,z\in E.$$
This is how one can prove that every Euclidian space satisfies (H).
Now consider the Euclidian space ${\mathbb R}^n$.

Does $E={\mathcal L}({\mathbb R}^n)$, endowed with the operator norm, satisfy (H) ? If so, is it $\ell^1$-embeddable ?

Edit. After Mateusz' answer, the two-dimensional case remains open.
 A: It does not hold, at least for $n \geq 3$. The reason is that $M_n$ contains then a subspace isometric to $\ell_{\infty}^3$ (diagonal matrices with three non-zero entries) and for this space the inequality does not hold -- take $x=(1,1,0)$, $y=(0,1,1)$ and $z=(1,0,1)$.
A: The inequality is true for $2 \times 2$ self-adjoint matrices, because of the following formula, which is reminiscent of $\max(|a|,|b|)=\frac{1}{2}|a+b|+\frac{1}{2}|a-b|$
$$ \|A\| = \frac{1}{2} | \mathrm{tr} A | + \frac{1}{\sqrt{2}} \| P(A) \|_2 $$
where $\|\cdot\|_2$ is the Hilbert-Schmidt (or Frobenius) norm and $P(A)= A - \frac{\mathrm{tr} A}{2} \mathrm{Id}$ is the orthogonal projection onto the subspace of trace $0$ matrices. Both terms in this sum satisfy Hlawka's inequality (the Hilbert--Schmidt norm is a Euclidean norm).
Geometrically the unit ball of $2 \times 2$ self-adjoint matrices looks like a tridimensional double-cone over a disk (extreme points are reflections and $\pm \mathrm{Id}$), which is a geometric way to interpret the formula above. In the non-self-adjoint case I don't have such a clear picture.
