Minkowski's first inequality

Remarks

One derivation uses the Brunn–Minkowski theorem. If $\operatorname{vol}(K) = 0$ the inequality is trivial, so assume $\operatorname{vol}(K) > 0$. The function $f(t) = \operatorname{vol}(K + tL)^{1/n}$ is concave on $[0, \infty)$, so $f'(0) \ge f(1) - f(0) \ge \operatorname{vol}(L)^{1/n}$. By the definition of mixed volumes, $\operatorname{vol}(K + tL) = \operatorname{vol}(K) + n V_1(K, L)\, t + O(t^2)$, so $f'(0) = V_1(K, L) \operatorname{vol}(K)^{(1-n)/n}$, which gives the inequality. It also follows from the Alexandrov–Fenchel inequality.

Example: take $L = B$, the closed unit ball, of volume $\kappa_n$. Then $n V_1(K, B)$ is the surface area $S(K)$, and the inequality becomes $S(K)^n \ge n^n \kappa_n \operatorname{vol}(K)^{n-1}$, the isoperimetric inequality for convex bodies. For $n = 2$ this is $\operatorname{per}(K)^2 \ge 4 \pi \operatorname{area}(K)$.

Going deeper

For a student

The inequality is a key step in the uniqueness part of the Minkowski problem: two convex bodies with the same surface area measure are translates of each other. It is the case $K_3 = \cdots = K_n = K$ of a chain of inequalities obtained from the Alexandrov–Fenchel inequality. Minkowski's second inequality, $V_1(K, L)^2 \ge \operatorname{vol}(K)\, V(K, \dots, K, L, L)$, is another case of that chain.