Gruber P. Convex and Discrete Geometry

6 Mixed Volumes and Quermassintegrals 101
By (12) the sum of the first two of these volumes equals the sum of the last two.
Equating the coefficients of λλ2 ···λd on both sides of this equality then yields the
theorem. ⊓⊔
Proof (using the continuity of mixed volumes and Lemma 6.5). Since the mixed
volumes are continuous in their entries by Theorem 6.8, it is sufficient to prove the
theorem in case where D2,...,Dd are polytopes.
We now show that, for the support functions of C ∪ D and C ∩ D, wehavethe
equalities,
hC∩D = min{hC, h D} and hC∪D = max{hC, h D}.
Let u ∈ E d \{o}. We may assume that hC(u) ≤ h D(u). Choose x ∈ C, y ∈ D such
that x · u = hC(u) and y · u = h D(u). The line segment from x to y is contained in
C ∪ D since C ∪ D is convex. Let z be the last point on this line segment contained
in C. Then z ∈ C ∩ D. Since
we have
and thus
C ∩ D ⊆ C ⊆{v : v · u ≤ x · u} ⊆{v : v · u ≤ z · u},
hC∩D(u) ≤ hC(u) ≤ z · u ≤ hC∩D(u)
hC∩D(u) = hC(u) = min{hC(u), h D(u)}.
This proves the first equality. The proof of the second equality is even simpler and
thus omitted.
These equalities yield the identity,
hC∪D + hC∩D = hC + h D.
Now use Lemma 6.5. ⊓⊔
Minkowski’s Inequalities
Mixed volumes satisfy several inequalities, in particular the following first and second
inequality of Minkowski [739].
Theorem 6.11. Let C, D ∈ C. Then:
(i) V (C, D,...,D) d ≥ V (C)V (D) d−1 , where for proper convex bodies C, D
equality holds if and only if C and D are (positive) homothetic.
(ii) V (C, D,...,D) 2 ≥ V (C, C, D,...,D)V (D).
The equality case in the second inequality is more complicated to formulate and to
prove. It was settled by Bol [139], thereby confirming a conjecture of Minkowski.
Proof. The Brunn–Minkowski theorem 8.3 shows that
(13) The function V � (1 − λ)C + λD � 1 d , 0 ≤ λ ≤ 1, is concave in λ. IfC
and D are proper, then this expression is linear if and only if C and D are
homothetic.