Reduction to convex case in Brunn - Minkowski Is there a direct way to reduct the general case of Brunn - Minkowski inequality $|A+B|^{1/n}\geqslant |A|^{1/n}+|B|^{1/n}$ for non-empty compact sets $A, B\subset \mathbb{R}^n$ (here $|\cdot|$ denotes Lebesgue measure in $\mathbb{R}^n$) to the case of convex compact sets?
 A: In a recent preprint of Conlon and Lim (link), they proved a discrete Brunn-Minkowski-type result  (Lemma 2.1) by using “compressions” to reduce to the case where $A,B$ are well-behaved connected subsets.
By using more aggressive “compressions” and induction, I think you should be able to reduce Brunn-Minkowski to the case where $A,B$ are both simplexes. Here’s a sketch:
Basically, given a subset $U$ and codimension-1 hyperplane $H$ with normal vector $v$, we define the $H$-compression of $U$, $U’ = C_H(U)$ , by making $U’ \cap (H+tv)$ a symmetric simplex based at $tv$ with $(d-1)$-dimensional volume equal to that of $U \cap (H+tv)$.
If you use $d$-hyperplanes $H_1,\dots,H_d$ whose normal vectors form an orthonormal basis, then repeatedly applying $C_{H_1},\dots, C_{H_d}$ to a set $U$ should converge towards a simplex with the same volume as $U$, which we denote $C(U)$.
We claim that $\mu(C(A)+C(B)) \le  \mu(A+B)$. It suffices to check that $\mu(C_H(A)+C_H(B)) \le  \mu(A+B)$ for any hyperplane $H$, this should follow from the assumption that Brunn-Minkowski holds on $(d-1)$-dimensions which implies $C_H(A)+C_H(B) \subset C_H(A+B)$. Thus, we’ve reduced to the case where $A,B$ are simplexes and thus convex.
