TITLE: When does Brunn-Minkowski inequatily $(m(A+B))^{1/d} \geqslant (m(A))^{1/d} + (m(B))^{1/d}$ become equality?
QUESTION [5 upvotes]: Let $A$ and $B$ be two non-empty compact subsets of $\mathbb{R}^d$, $m$ means Lebesgue measure.
Brunn-Minkowski inequality gives
$(m(A+B))^{1/d} \geqslant (m(A))^{1/d} + (m(B))^{1/d}$.
But how to prove the following?

$(m(A+B))^{1/d} = (m(A))^{1/d} + (m(B))^{1/d} \implies$
$A$ and $B$ are convex and $\exists$ $\delta＞0$, $h\in \mathbb{R}^d$ s.t. $A = \delta B + h$

(It's Problem $8$ of Chapter $1$, Page $48$ in E.M.Stein's Real Analysis  , though this problem doen't have much to do with real analysis.)
Thanks in advance.

REPLY [3 votes]: You might also be interested in the following two papers by Fields medalist A. Figalli:

Quantitative stability for the Brunn-Minkowski inequality,
Stability result for the Brunn-Minkowski inequality.

The stability result, proven with optimal transport techniques, implies in particular the equality condition.