[proofplan] The proof is a direct application of convexity of $K$ followed by the [Brunn-Minkowski inequality](/theorems/4080) in the fibre variable. First we show that convex combinations of points in two fibres lie in the fibre over the corresponding convex combination of base points. This gives a Minkowski-sum inclusion of fibres. Applying Brunn-Minkowski in $\mathbb{R}^m$ to that inclusion yields concavity of the $m$-th root of the slice volume, and the log-concavity statement follows from the weighted arithmetic-geometric mean inequality. [/proofplan]

[step:Verify that the projection is convex and the fibres are measurable compact sets] Define the coordinate projection map \begin{align*} \pi: K &\to \mathbb{R}^n \\ (x,y) &\mapsto x, \end{align*} and let $P := \pi(K)$ denote the projection of $K$ onto $\mathbb{R}^n$. For each $x \in \mathbb{R}^n$, define the fibre $K_x \subset \mathbb{R}^m$ as in the theorem statement. We first prove that $K_x$ is compact. Since $K$ is a convex body, $K$ is compact in $\mathbb{R}^{n+m}$, hence closed and bounded. If $K_x = \varnothing$, then $K_x$ is compact. If $K_x \neq \varnothing$, boundedness follows from boundedness of $K$: there exists $R > 0$ such that $K \subset \overline{B}(0,R) \subset \mathbb{R}^{n+m}$, and therefore $|y| \leq R$ for every $y \in K_x$. To prove closedness, let $(y_j)_{j=1}^{\infty}$ be a sequence in $K_x$ converging to $y \in \mathbb{R}^m$. Then $(x,y_j) \in K$ for every $j$, and $(x,y_j) \to (x,y)$ in $\mathbb{R}^{n+m}$. Since $K$ is closed, $(x,y) \in K$, so $y \in K_x$. Thus $K_x$ is closed and bounded in $\mathbb{R}^m$, and the [Heine-Borel Theorem](/theorems/???) implies that $K_x$ is compact. Convexity follows because if $y_0,y_1 \in K_x$ and $s \in [0,1]$, then $(x,y_0),(x,y_1) \in K$, so \begin{align*} (1-s)(x,y_0)+s(x,y_1) = (x,(1-s)y_0+sy_1) \in K, \end{align*} which gives $(1-s)y_0+sy_1 \in K_x$. Since compact subsets of $\mathbb{R}^m$ are Borel measurable, $\mathcal{L}^m(K_x)$ is well-defined for every $x \in \mathbb{R}^n$. The projection $P$ is convex: if $x_0,x_1 \in P$, choose $y_0 \in K_{x_0}$ and $y_1 \in K_{x_1}$. For $t \in [0,1]$, convexity of $K$ gives \begin{align*} (1-t)(x_0,y_0)+t(x_1,y_1) = ((1-t)x_0+tx_1,(1-t)y_0+ty_1) \in K, \end{align*} so $(1-t)x_0+tx_1 \in P$. [/step]

[step:Use convexity of $K$ to compare Minkowski sums of fibres] Fix $x_0,x_1 \in P$ and $t \in [0,1]$. Define the Minkowski combination \begin{align*} (1-t)K_{x_0}+tK_{x_1} := \{(1-t)y_0+ty_1 : y_0 \in K_{x_0},\ y_1 \in K_{x_1}\} \subset \mathbb{R}^m. \end{align*} We claim that \begin{align*} (1-t)K_{x_0}+tK_{x_1} \subset K_{(1-t)x_0+tx_1}. \end{align*} Let $y \in (1-t)K_{x_0}+tK_{x_1}$. By definition of this Minkowski combination, there exist $y_0 \in K_{x_0}$ and $y_1 \in K_{x_1}$ such that \begin{align*} y=(1-t)y_0+ty_1. \end{align*} Since $y_0 \in K_{x_0}$ and $y_1 \in K_{x_1}$, we have $(x_0,y_0),(x_1,y_1) \in K$. Convexity of $K$ gives \begin{align*} (1-t)(x_0,y_0)+t(x_1,y_1) = ((1-t)x_0+tx_1,y) \in K. \end{align*} Therefore $y \in K_{(1-t)x_0+tx_1}$, proving the inclusion. [/step]

[step:Apply Brunn-Minkowski in the fibre space]Because $x_0,x_1 \in P$, the fibres $K_{x_0}$ and $K_{x_1}$ are nonempty. By the first step, they are compact Borel subsets of $\mathbb{R}^m$. Since $m \in \mathbb{N}$, we have $m \geq 1$, so the $m$-dimensional [Brunn-Minkowski Inequality](/theorems/???) applies to these measurable subsets of $\mathbb{R}^m$. Applied to $K_{x_0}$ and $K_{x_1}$ with weights $1-t$ and $t$, it gives \begin{align*} \mathcal{L}^m((1-t)K_{x_0}+tK_{x_1})^{1/m} \geq (1-t)\mathcal{L}^m(K_{x_0})^{1/m} + t\mathcal{L}^m(K_{x_1})^{1/m}. \end{align*} By the fibre inclusion proved above and monotonicity of Lebesgue measure, \begin{align*} \mathcal{L}^m(K_{(1-t)x_0+tx_1}) \geq \mathcal{L}^m((1-t)K_{x_0}+tK_{x_1}). \end{align*} Since the function $r \mapsto r^{1/m}$ is increasing on $[0,\infty)$, we obtain \begin{align*} \mathcal{L}^m(K_{(1-t)x_0+tx_1})^{1/m} \geq \mathcal{L}^m((1-t)K_{x_0}+tK_{x_1})^{1/m}. \end{align*} Combining the last two inequalities yields \begin{align*} \mathcal{L}^m(K_{(1-t)x_0+tx_1})^{1/m} \geq (1-t)\mathcal{L}^m(K_{x_0})^{1/m} + t\mathcal{L}^m(K_{x_1})^{1/m}. \end{align*} Define \begin{align*} f: P &\to [0,\infty) \\ x &\mapsto \mathcal{L}^m(K_x)^{1/m}. \end{align*} The last inequality is precisely the concavity of $f$ on $P$.[/step]

[guided]We now use the only external ingredient in the argument: the [Brunn-Minkowski Inequality](/theorems/???) in the fibre space $\mathbb{R}^m$. We verify its hypotheses directly. Since $x_0,x_1 \in P$, the definition of $P$ gives points $y_0 \in K_{x_0}$ and $y_1 \in K_{x_1}$, so both fibres are nonempty. Each fibre is a compact Borel subset of $\mathbb{R}^m$: compactness follows because it is closed and bounded as a fibre of the compact set $K \subset \mathbb{R}^{n+m}$, and Borel measurability follows because compact subsets of $\mathbb{R}^m$ are closed. Finally, $m \in \mathbb{N}$, so $m \geq 1$ and the $m$-dimensional inequality is applicable. Applying Brunn-Minkowski to $K_{x_0}$ and $K_{x_1}$ with weights $1-t$ and $t$ gives \begin{align*} \mathcal{L}^m((1-t)K_{x_0}+tK_{x_1})^{1/m} \geq (1-t)\mathcal{L}^m(K_{x_0})^{1/m} + t\mathcal{L}^m(K_{x_1})^{1/m}. \end{align*} This estimate is exactly about the Minkowski combination of the two fibres. The convexity step showed that this Minkowski combination is contained in the fibre over the convex combination of the base points: \begin{align*} (1-t)K_{x_0}+tK_{x_1} \subset K_{(1-t)x_0+tx_1}. \end{align*} Lebesgue measure is monotone on measurable sets, so this inclusion implies \begin{align*} \mathcal{L}^m(K_{(1-t)x_0+tx_1}) \geq \mathcal{L}^m((1-t)K_{x_0}+tK_{x_1}). \end{align*} Because $r \mapsto r^{1/m}$ is increasing on $[0,\infty)$, taking $m$-th roots preserves the inequality: \begin{align*} \mathcal{L}^m(K_{(1-t)x_0+tx_1})^{1/m} \geq \mathcal{L}^m((1-t)K_{x_0}+tK_{x_1})^{1/m}. \end{align*} Combining this monotonicity estimate with Brunn-Minkowski gives \begin{align*} \mathcal{L}^m(K_{(1-t)x_0+tx_1})^{1/m} \geq (1-t)\mathcal{L}^m(K_{x_0})^{1/m} + t\mathcal{L}^m(K_{x_1})^{1/m}. \end{align*} This is the desired concavity inequality for the function $f(x)=\mathcal{L}^m(K_x)^{1/m}$.[/guided]

[step:Derive log-concavity of the slice-volume function] Define the slice-volume function \begin{align*} g: P &\to [0,\infty) \\ x &\mapsto \mathcal{L}^m(K_x). \end{align*} The concavity just proved says \begin{align*} g((1-t)x_0+tx_1)^{1/m} \geq (1-t)g(x_0)^{1/m}+t g(x_1)^{1/m}. \end{align*} If $g(x_0)=0$ or $g(x_1)=0$, then log-concavity in the extended nonnegative sense follows because the right-hand side $g(x_0)^{1-t}g(x_1)^t$ is $0$ for $t \in (0,1)$, while $g$ is nonnegative. If $g(x_0)>0$ and $g(x_1)>0$, the weighted arithmetic-geometric mean inequality gives \begin{align*} (1-t)g(x_0)^{1/m}+t g(x_1)^{1/m} \geq g(x_0)^{(1-t)/m}g(x_1)^{t/m}. \end{align*} Therefore \begin{align*} g((1-t)x_0+tx_1)^{1/m} \geq g(x_0)^{(1-t)/m}g(x_1)^{t/m}. \end{align*} Raising both sides to the power $m$ gives \begin{align*} g((1-t)x_0+tx_1) \geq g(x_0)^{1-t}g(x_1)^t. \end{align*} Thus $g$ is log-concave on $P$. This completes the proof. [/step]