= Brunn-Minkowski theorem
{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 inequality
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{synonym}
Let $n\geq1$ and let $A,B\subseteq\mathbb R^n$ be nonempty <Lebesgue measurable sets> with finite <Lebesgue measures>. Assume that $A+B$ is also a <Lebesgue measurable set>; its <measure> may be infinite. Then
$$
\lambda_n(A+B)^{1/n}\geq\lambda_n(A)^{1/n}+\lambda_n(B)^{1/n}.
$$
Here $\lambda_n$ denotes $n$-dimensional <Lebesgue measure> in <Euclidean space>. Nonempty <compact sets> and nonempty <open sets> of finite <measure> satisfy the measurability assumptions. Measurability of arbitrary summands alone does not guarantee measurability of their <Minkowski sum>. The theorem does not require <convexity>. See https://sites.stat.washington.edu/jaw/COURSES/520s/523/HO.523.20/Gardner.BullAMS.02.pdf[R. J. Gardner, The Brunn-Minkowski Inequality (2002), Theorem 4.1 and the following remarks].
The $n$th root converts a volume into a quantity that scales like a <length>: <uniform dilation> gives $\lambda_n(tA)^{1/n}=t\lambda_n(A)^{1/n}$ for $t>0$. In the <Euclidean plane> the theorem compares square roots of <areas>, not the <areas> themselves. For the two <rectangles> above it reads $3\geq\sqrt2+\sqrt2=2\sqrt2$. Equivalently their sum has <area> $9\geq8$. For the two <disks>, it reads $3\sqrt\pi=\sqrt\pi+2\sqrt\pi$, an equality. These are exact geometric examples, not measured data.
Applying the theorem to the scaled <sets> gives, whenever that sum is a <Lebesgue measurable set>,
$$
\lambda_n((1-t)A+tB)^{1/n}\geq(1-t)\lambda_n(A)^{1/n}+t\lambda_n(B)^{1/n},\qquad 0<t<1.
$$
For <convex bodies>, this means the root-volume of the interpolated shape lies above the straight line between the input root-volumes. In fact $t\mapsto\lambda_n((1-t)A+tB)^{1/n}$ is a <concave function> on $[0,1]$: <convexity> gives $sA+tA=(s+t)A$ for $s,t\geq0$, so interpolating two intermediate shapes gives another intermediate shape and the same inequality applies. The beauty is that no particular boundary shape appears in the bound: possible pointwise displacements force a minimum amount of space.
For <convex bodies> $A,B$, equality holds exactly when they are <homothetic convex bodies>, meaning $B=v+sA$ for some <vector> $v$ and $s>0$. This equality qualification is part of the <Brunn-Minkowski theorem> for <convex bodies>. https://sites.stat.washington.edu/jaw/COURSES/520s/523/HO.523.20/Gardner.BullAMS.02.pdf[Gardner (2002), introduction, equation 2 and its equality condition]{ref} For positive-sided axis-aligned <rectangles>, the comparison can be checked directly:
$$
(a+c)(b+d)-(\sqrt{ab}+\sqrt{cd})^2=(\sqrt{ad}-\sqrt{bc})^2\geq0.
$$
Equality requires $ad=bc$: the <rectangles> must have the same aspect ratio. Our $2\times1$ and $1\times2$ <rectangles> do not. All positive-radius <disks> are <homothetic convex bodies>, explaining their equality.
Do not apply that equality criterion without its hypotheses. If $A=\{a\}$, its <Lebesgue measure> is zero and $A+B$ is a <translation> of $B$, so equality holds even for nonconvex $B$. Also, changing a summand on a <null set> can change the <measure> of its <Minkowski sum>: in one dimension, $A=[0,1]\cup\{3\}$ and $B=[0,1]$ both have <measure> $1$, but $A+B=[0,2]\cup[3,4]$ has <measure> $3$. Removing the single extra point restores equality with sum <measure> $2$. Thus merely being equal almost everywhere to <homothetic convex bodies> is not a sufficient equality test for arbitrary <sets>.
The <Brunn-Minkowski theorem> also bounds the collision-region <volume>: for positive-volume <compact sets> $O,R$, reflection preserves <Lebesgue measure>, so $\lambda_n(O+(-R))^{1/n}\geq\lambda_n(O)^{1/n}+\lambda_n(R)^{1/n}$. In the toy collision example this gives <area> at least $(2+\sqrt2)^2=6+4\sqrt2\approx11.657$, compared with the exact <area> $12$. A lower bound alone does not identify safe reference <positions>; the actual <Minkowski sum> supplies those.
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