Brunn-Minkowski, Mixed Volumes & the Isoperimetric Inequality

Adding shapes: the Minkowski sum

In the previous guides you met a convex body K as a compact convex set with nonempty interior, pinned down by its supporting hyperplanes and reconstructed from its extreme points by Krein-Milman. Now we let bodies interact. Given two convex bodies K and L in R^n, their Minkowski sum is K + L = { x + y : x in K, y in L }. Concretely, slide a copy of L so its origin sits at each point of K and sweep; the union is K + L, which is again convex. A clean way to track it is the support function h_K(u) = max over x in K of <x, u>: it is simply additive under the sum, h_{K+L} = h_K + h_L.

Scaling enters too: for a real number t >= 0 the dilate tK = { t x : x in K } has h_{tK} = t h_K. So expressions like K + tL describe a body growing in the direction of L. The headline example is L = B, the unit ball: K + tB is exactly the set of points within distance t of K, the outer parallel body or 'r-neighborhood' of K. Watching the volume of K + tB grow as t increases is the doorway to surface area, and ultimately to the isoperimetric inequality.

Volume becomes a polynomial: mixed volumes

Here is the structural miracle that organizes the whole subject. Fix convex bodies K_1, ..., K_m and nonnegative reals t_1, ..., t_m. Then the volume of the combination, V(t_1 K_1 + ... + t_m K_m), is a homogeneous polynomial of degree n in the variables t_1, ..., t_m. This is Minkowski's theorem on mixed volumes. The coefficients of that polynomial, suitably normalized and symmetrized, are the mixed volumes V(K_{i_1}, ..., K_{i_n}), one number for each unordered choice of n bodies (with repetition) from your list.

The two-body case is the one to internalize. Write V(K + tL) as a polynomial in t for bodies in R^n. The constant term is vol(K), the top term is t^n vol(L), and the intermediate coefficients are the mixed volumes V(K, ..., K, L, ..., L) counting how many copies of each appear. In particular the very first variation coefficient, the one multiplying t to the first power, equals n times the mixed volume V(K, ..., K, L) with n-1 copies of K and one L.

V(K + t L) = sum_{j=0}^{n} C(n,j) * W_j(K,L) * t^j      (binomial expansion)

  n = 2:  V(K + t L) = vol(K) + 2 V(K,L) t + vol(L) t^2
  n = 3:  V(K + t L) = vol(K) + 3 V(K,K,L) t + 3 V(K,L,L) t^2 + vol(L) t^3

  L = B (unit ball):  d/dt V(K + tB) at t=0  =  surface area S(K)

Why care? Because surface area, the object of the isoperimetric problem, is now a mixed volume. Taking L = B, the relation S(K) = n V(K, ..., K, B) (with n-1 copies of K) recovers the classical surface area as a derivative of volume, the Steiner formula in disguise. Mixed volumes thus unify volume, surface area, mean width, and intrinsic volumes into a single algebraic gadget. They are also monotone and translation-invariant, which is exactly the toolkit the next inequalities need.

The Brunn-Minkowski inequality

Now the centerpiece. The Brunn-Minkowski inequality says that for nonempty compact sets (in particular convex bodies) K and L in R^n, the n-th root of volume is concave under Minkowski addition: vol(K + L)^(1/n) >= vol(K)^(1/n) + vol(L)^(1/n). Equality holds exactly when K and L are homothetic — the same shape up to translation and positive scaling. Read it as: the 'linear-dimension content' vol^(1/n) is superadditive, never less than the sum of the pieces.

A one-line sanity check in dimension 1: intervals of lengths a and b add to an interval of length a + b, and indeed (a + b)^1 = a^1 + b^1 with equality — intervals are always homothetic, so equality is forced. In dimension 2, take a unit square K and a unit disk L. Their sum is the rounded square of the earlier callout, and the inequality says its area's square root is at least 1 + sqrt(pi) — true, because the rounded square comfortably contains both a translate of K and a translate of L plus the connecting strips.

There is an equivalent dimension-free form that is often more useful: for t in [0, 1], vol((1-t)K + tL)^(1/n) >= (1-t) vol(K)^(1/n) + t vol(L)^(1/n). This is literally the statement that t -> vol((1-t)K + tL)^(1/n) is a concave function on [0, 1]. The slice picture explains why: if you cut a convex body in R^n by parallel hyperplanes, the (n-1)-volume of the slice, raised to the power 1/(n-1), varies concavely with the height — Brunn's theorem. Brunn-Minkowski is that slice concavity, packaged.

From concavity to the isoperimetric inequality

Now the payoff that gives this guide its title. The isoperimetric inequality says: among all bodies of a given surface area, the ball encloses the most volume; equivalently, among all bodies of given volume, the ball has the least surface area. The sharp form in R^n reads S(K)^n >= n^n omega_n vol(K)^(n-1), where omega_n = vol(B) is the volume of the unit ball, with equality if and only if K is a ball. The slogan 'round is best' is finally a theorem with hypotheses.

1. Start from Brunn-Minkowski applied to K and the unit ball B: vol(K + tB)^(1/n) >= vol(K)^(1/n) + t vol(B)^(1/n) for t >= 0.
2. Raise both sides to the n-th power and expand the right side by the binomial theorem; keep terms up to first order in t.
3. Subtract vol(K) from both sides and divide by t, then let t -> 0+. The left side becomes the derivative d/dt vol(K + tB) at 0, which is the surface area S(K).
4. The first-order term on the right is n vol(K)^((n-1)/n) vol(B)^(1/n). Rearranging S(K) >= n vol(K)^((n-1)/n) omega_n^(1/n) and raising to the n-th power gives the sharp isoperimetric inequality.

That derivation is the cleanest reason mixed volumes deserve their billing: surface area S(K) is literally the t-derivative of vol(K + tB), and Brunn-Minkowski controls that derivative from below by a pure power of vol(K). The equality case carries over too — equality in Brunn-Minkowski needs K homothetic to B, which forces K itself to be a ball. So the same homothety condition that looked abstract three paragraphs ago is exactly what crowns the sphere.

Going further: Alexandrov-Fenchel and where it bites

Brunn-Minkowski is the n = 2 shadow of a far deeper statement about mixed volumes. The Alexandrov-Fenchel inequality says that for convex bodies K_1, ..., K_n, the mixed volume satisfies V(K_1, K_2, K_3, ..., K_n)^2 >= V(K_1, K_1, K_3, ..., K_n) V(K_2, K_2, K_3, ..., K_n). Read with the last n-2 bodies held fixed, this is a log-concavity statement: the map K -> V(K, K, fixed...) behaves like a concave quadratic form. Specializing every fixed slot to the same body recovers Brunn-Minkowski; specializing to balls recovers a whole ladder of classical inequalities among the intrinsic volumes.

Be honest about difficulty here, in the spirit of the rest of this ladder. Alexandrov-Fenchel is genuinely hard: the original proofs are intricate, the equality cases were not fully resolved until very recent work, and there is no short slick argument of the Brunn-Minkowski caliber. We state and motivate it; a real proof is a substantial chapter, not a paragraph. Treat its appearance here as a signpost to the literature, not a result we have established.

Two warnings before you move on. First, Brunn-Minkowski as stated needs the sets to be nonempty and (for the equality characterization) convex; for non-convex measurable sets the inequality still holds via the Prekopa-Leindler approach, but the homothety equality case can fail. Second, mixed volumes can be zero or behave degenerately when bodies are lower-dimensional, so 'V(K, L) > 0' is an assumption, not a freebie. The machinery you have just seen — Minkowski sums, the volume polynomial, the 1/n-concavity — is also exactly the convex-geometry input behind Minkowski's lattice theorem in the next guide, where volume bounds force a symmetric convex body to swallow a lattice point.