Suppose $f: \mathbb{R}^d \to \mathbb{R}$ is a smooth convex function.
Consider the level sets of the function, namely $M_s = \{x: f(x) = s\}$. Is it true/known that the surface areas of $M_s$ are log-concave as a function of $s$?
(This feels awfully like a Brunn-Minkowski style inequality, but I'm unsure if it follows from known results. If so, references are highly appreciated!)
Yes, this is true, and you are right, this follows from a generalization of the Brunn-Minkowski inequality.
Let $K_s = \{x \mid f(x) \le s\}$, so that $M_s = \partial K_s$. We have $K_s \supseteq (1-s)K_0 + sK_1$, thus the surface area of the former is $\ge$ the surface area of the latter.
The surface area of a convex body can be written as a mixed volume: $$ \mathrm{vol}_{n-1}(\partial K) = n \mathrm{vol}(\underbrace{K, \ldots, K}_{n-1}, B). $$ A general version of the Brunn-Minkowski inequality says that the function $$ \mathrm{vol}(\underbrace{(1-t)K_0 + tK_1, \ldots, (1-t)K_0 + tK_1}_{n-i}, L_1, \ldots, L_i)^{\frac1{n-i}} $$ is concave for any convex bodies $L_1, \ldots, L_i$.
It follows that the $(n-1)$-st root of the surface area of $K_t$ is concave. Concavity of $f^{\frac{1}{n-1}}$ implies concavity of $\log f$, and we are done.
References:
Gardner, R. J., The Brunn-Minkowski inequality, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008, Section 17.
Burago, Yu. D.; Zalgaller, V. A., Geometric inequalities. Transl. from the Russian by A. B. Sossinsky, Grundlehren der Mathematischen Wissenschaften, 285. Berlin etc.: Springer-Verlag. XIV, 331 p.; DM 184,- (1988). ZBL0633.53002, p. 146.
Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001, Theorem 6.4.3.
