the Brunn-Minkowski inequality
/ BROON min-KOF-skee /
Take two solid shapes and 'add' them by sliding one over every point of the other and sweeping out all the sums — the Minkowski sum. Brunn-Minkowski says the combined shape is always at least as voluminous as you would naively expect: the n-th root of its volume is at least the sum of the n-th roots of the two original volumes. In a single inequality it captures that mixing convex shapes never wastes room — convexity makes volume behave super-additively under this sum.
Precisely, for nonempty compact sets A and B in R^n, the Minkowski sum is A + B = {a + b : a in A, b in B}, and the inequality states vol(A + B)^{1/n} >= vol(A)^{1/n} + vol(B)^{1/n}. Equivalently, for t in [0, 1], vol(t*A + (1-t)*B)^{1/n} >= t*vol(A)^{1/n} + (1-t)*vol(B)^{1/n}: the n-th root of volume is a concave function along Minkowski interpolation. Equality holds (for convex bodies) exactly when A and B are homothetic — the same shape up to translation and scaling. One clean way to see it for boxes: the sum of two axis-aligned boxes is a box whose side lengths add, and the inequality on side lengths follows from the AM-GM inequality; the general case is a limiting/rearrangement argument.
Brunn-Minkowski is one of the most consequential inequalities in mathematics. The classical isoperimetric inequality — among all bodies of a given volume the ball has the least surface area — drops out by taking B a tiny ball and differentiating. It anchors the theory of mixed volumes and the Alexandrov-Fenchel inequality, it gives concentration-of-measure and dimension-free estimates in probability and analysis, and its functional cousin is the Prekopa-Leindler inequality. Two honest cautions: the 1/n power is essential — plain volume is NOT super-additive, vol(A+B) >= vol(A) + vol(B) is false in general — and the sharp equality case requires convexity; for non-convex sets the inequality still holds but equality can occur in degenerate ways.
Let A and B both be the unit cube [0,1]^3 in R^3. Their Minkowski sum is the cube [0,2]^3 of volume 8. Check Brunn-Minkowski: vol(A+B)^{1/3} = 8^{1/3} = 2, while vol(A)^{1/3} + vol(B)^{1/3} = 1 + 1 = 2. Equality holds — exactly as predicted, since A and B are homothetic (in fact identical). Plain volumes would have suggested 8 >= 1 + 1, true but far weaker; the cube-root form is the sharp statement.
Two unit cubes sum to a 2x2x2 cube; the 1/3-power form holds with equality (homothetic case).
The dimension-1/n power cannot be dropped: 'vol is super-additive' is plainly false (a thin slab plus a perpendicular thin slab has tiny pieces but their sum can be large, and degenerate flat sets have volume 0 yet positive-volume sums). Only the 1/n-th root makes the clean concavity true.