Recall that a Hilbert geometry is the interior of a convex body $K \subset \mathbb{R}^n$ provided with the metric $$ d(x,y) = \frac{1}{2} \ln\left(\frac{|x-b|}{|y-b|}\frac{|y-a|}{|x-a|}\right) , $$ where $a$ and $b$ are the points of intersection of the boundary of $K$ and the straight line determined by $x$ and $y$ with the provision that $x$ lie in the segment $ay$ (and $y$ lie in the segment $xb$).

Question. Given two metric balls $B_1$ and $B_2$ Hausdorff measure $\nu > 0$ in a Hilbert geometry $(K,d_K)$, is it true that the Hausdorff measure of $(B_1 + B_2)/2$, defined as the set formed by the midpoints of all line segments having one endpoint in $B_1$ and another endpoint in $B_2$, is never greater than $\nu$?

Motivation. Given two subsets $A$ and $B$ of a geodesic metric space $(X,d)$, let $(A + B)/2$ denote the set of midpoints of all minimal geodesic segments with one endpoint in $A$ and another endpoint in $B$.

Given a Borel measure $\mu$ on $X$, let us say that the metric-measure space $(X,d,\mu)$ is a Brunn-Minkowski space if whenever $A$ and $B$ are open subsets of $X$ with finite measure, then $$ \mu \left( \frac{A + B}{2}\right) \geq \sqrt{\mu(A)\mu(B)} . $$

Loosely speaking, this is the sort of inequality that after the work of Cordero-Erausquin, McCann and Schmuckenschläger, Sturm, and Lott and Villani (and many others) has come to be associated with non-negative Ricci curvature. In fact, it is the weakest possible related notion I could come up with. The condition in the OP aims to check for "non-positive" Ricci curvature.

Hilbert geometries generalize hyperbolic geometry (although they could also be flat, as when $K$ is a simplex), but it is very hard to come up with a good, geometric way to say "they are negatively curved". In many cases they are indeed Gromov hyperbolic (yves Benoist wrote a fantastically pretty paper on this), but it is also known that if a Hilbert geometry is on-positively curved in the sense of Busemann, then it is hyperbolic space. I'm just trying to come up with something that seems reasonable making use that Hilbert geometries + Hausdorff measure (or Holmes-Thompson volume) are nice metric-measure spaces.

Remark. Finsler spaces can be very different from Riemannian spaces. For example, compact, convex surfaces in normed spaces are not necessarily Brunn-Minkowski spaces (and their Ricci curvature in the Sturm-Lott-Villani sense is not non-negative). C. Vernicos and I found a cute little example of this phenomenon.