= Brunn–Minkowski inequality
{c}
{title2=$\lambda_n(A+B)^{1/n}\geq\lambda_n(A)^{1/n}+\lambda_n(B)^{1/n}$}
{wiki=Brunn–Minkowski_theorem}
= Brunn–Minkowski theorem
{c}
{synonym}
For nonempty <open sets> $A,B\subset\mathbb R^n$, the <Lebesgue measure> of their <Minkowski sum> satisfies
$$
\lambda_n(A+B)^{1/n}\geq\lambda_n(A)^{1/n}+\lambda_n(B)^{1/n}.
$$
It also holds for compact sets and in standard measurable-set formulations with the appropriate measurability qualification. Normalize both volumes to one and apply the <Prékopa–Leindler inequality> to <indicator functions>.
Back to article page
