I'm reading something on Brunn-Minkowski inequality and I've come along the following fact: the inequality, which says that if K and L are convex bodies then $$V((1-\lambda)K+\lambda L)^{1/n}\geq (1-\lambda)V(K)^{1/n}+\lambda V(L)^{1/n}$$ implies that the function $$f_1(t)=V((1-t)K+tL)^{1/n}$$ is concave in [0,1]. Why? I can see the reason why $f_2(K)=V(K)^{1/n}$ is convex in the space of convex bodies equipped with Minkowski sum, but the concavity of $f_1:[0,1]\rightarrow \mathbb{R}$ is not clear to me.
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$\begingroup$ Try to proof the definition of a concave function (see e.g. wikipedia) using the Brunn-Minkowski inequality. This should work. Your second function as it is defined is constant (it is not mapping convex bodies) $\endgroup$– sqlmanJun 12, 2021 at 10:27
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$\begingroup$ edited so that $f_2$ isn't constant $\endgroup$– toyr99Jun 12, 2021 at 10:42
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