Maximal intersection of slabs in $\mathbb{R}^n$ with a compact convex centrally symmetric set Let $K \in \mathbb{R}^n$ be a compact convex set containing the origin and symmetric with respect to the origin.
Let $S_i(t_i)$ be a finite set of slabs of various widths and orientations, translated from the origin by distance $t_i$.
Consider the set $I = \left(\bigcap_iS_i(t_i) \right)\cap K$.
How do I show that $I$ has maximum volume in $\mathbb{R}^n$ when $t_i = 0$ for all $i$,  i.e. all slabs are centered at the origin?
Notes

*

*A slab centered at $0$ is a set of the form $\{x:|\phi(x)|\le 1\}$ where $\phi$ is some linear functional. In the problem the slabs may be translated away from $0$.


*The one-dimensional case $n=1$ is very easy, because putting all slabs (intervals) so that they have a common center amounts to having just one of them, the smallest. And since $K$ is also an interval, the conclusion follows. However, the case $n\ge 2$ is nothing like this.
 A: For any convex sets $K$ and $L$, the map $\mathbb R^n\to\mathbb R$, $v\mapsto \operatorname{vol}(K\cap (L+v))^{1/n}$ is concave on its support, which is $K-L$.  (See below.)  If $K$ and $L$ are both origin-symmetric, then $K\cap (L+v)$ and $K\cap (L-v)$ are congruent, so they have the same volume, whence
\begin{align*}
\operatorname{vol}(K\cap L)^{1/n}
&\ge \tfrac12 \operatorname{vol}(K\cap(L+v))^{1/n} + \tfrac12 \operatorname{vol}(K\cap (L-v))^{1/n} \\
&= \operatorname{vol}(K\cap(L+v))^{1/n}
\end{align*}
as desired.  (Replace $K,L,v$ with $K\cap\bigcap_{i\ne j} (S_i+t_i),S_j,t_j$ and proceed by induction.)
To prove the concavity, consider the set
$$ \{(x,y)\in\mathbb R^n\times\mathbb R^n : x\in K, y\in\mathbb R^n\}
\cap \{(x,y)\in\mathbb R^n\times\mathbb R^n : x\in\mathbb R^n, y\in L+x\}
$$
This set is convex (because it's an intersection of two cylinders, one right and one oblique, which are convex), and its sections by flats of the form $\{(x,-v) : x\in\mathbb R^n\}$ are (congruent to) the sets $K\cap(L+v)$.  Thus the $n$th root of the volumes of these sections is a concave function, by Brunn's theorem (a version of the Brunn-Minkowski inequality).
This concavity result is well-known in convexity; it's from Fáry and Rédei, "Der zentralsymmetrische Kern und die zentralsymmetrische Hülle von convexen Korpern", Math. Ann. 122 (1950), 205–220.
