[proofplan] We prove the full concavity inequality directly. Fix two interpolation parameters $s,t \in [0,1]$ and a mixing parameter $\lambda \in [0,1]$, then identify the Minkowski interpolation between $K_s$ and $K_t$ with the single interpolation set $K_{(1-\lambda)s+\lambda t}$. Applying the [Brunn-Minkowski inequality](/theorems/4080) to $(1-\lambda)K_s$ and $\lambda K_t$, and using the homogeneity of Lebesgue measure under scalar dilation, gives precisely the desired concavity inequality. [/proofplan] [step:Identify the interpolation between $K_s$ and $K_t$ as a single Minkowski interpolation] Fix $s,t,\lambda \in [0,1]$, and define \begin{align*} r := (1-\lambda)s+\lambda t. \end{align*} For each $u \in [0,1]$, recall that \begin{align*} K_u = (1-u)K + uL. \end{align*} We claim that \begin{align*} (1-\lambda)K_s+\lambda K_t = K_r. \end{align*} Indeed, \begin{align*} (1-\lambda)K_s+\lambda K_t &= (1-\lambda)\big((1-s)K+sL\big)+\lambda\big((1-t)K+tL\big) \\ &= \big((1-\lambda)(1-s)+\lambda(1-t)\big)K +\big((1-\lambda)s+\lambda t\big)L \\ &= (1-r)K+rL \\ &= K_r. \end{align*} The second equality uses convexity of $K$ and $L$: for any nonnegative scalars $a,b$ and any convex set $C \subset \mathbb{R}^n$, \begin{align*} aC+bC=(a+b)C. \end{align*} If $a+b=0$, both sides equal $\{0\}$. If $a+b>0$, then \begin{align*} ax+by=(a+b)\left(\frac{a}{a+b}x+\frac{b}{a+b}y\right) \in (a+b)C \end{align*} for $x,y \in C$, while the reverse inclusion follows by writing $(a+b)z=az+bz$ for $z \in C$. [guided] Fix $s,t,\lambda \in [0,1]$, and define the target interpolation parameter \begin{align*} r := (1-\lambda)s+\lambda t. \end{align*} The goal is to compare the volume radius at $r$ with the volume radii at $s$ and $t$. To do this through Brunn-Minkowski, we must first rewrite the set $K_r$ as a Minkowski sum involving $K_s$ and $K_t$. For each $u \in [0,1]$, the interpolation set is \begin{align*} K_u = (1-u)K+uL. \end{align*} We compute: \begin{align*} (1-\lambda)K_s+\lambda K_t &= (1-\lambda)\big((1-s)K+sL\big)+\lambda\big((1-t)K+tL\big) \\ &= \big((1-\lambda)(1-s)+\lambda(1-t)\big)K +\big((1-\lambda)s+\lambda t\big)L. \end{align*} The coefficient of $L$ is exactly $r$, and the coefficient of $K$ is \begin{align*} (1-\lambda)(1-s)+\lambda(1-t) &= (1-\lambda)- (1-\lambda)s+\lambda-\lambda t \\ &= 1-\big((1-\lambda)s+\lambda t\big) \\ &= 1-r. \end{align*} Therefore \begin{align*} (1-\lambda)K_s+\lambda K_t=(1-r)K+rL=K_r. \end{align*} The only set-theoretic point hidden in the algebra is the rule $aC+bC=(a+b)C$ for a convex set $C \subset \mathbb{R}^n$ and nonnegative scalars $a,b$. If $a+b=0$, then $a=b=0$ and both sides are $\{0\}$. If $a+b>0$, then every element of $aC+bC$ has the form $ax+by$ with $x,y \in C$, and \begin{align*} ax+by=(a+b)\left(\frac{a}{a+b}x+\frac{b}{a+b}y\right). \end{align*} Since $C$ is convex, the point inside the parentheses belongs to $C$, so $ax+by \in (a+b)C$. Conversely, for every $z \in C$, \begin{align*} (a+b)z=az+bz \in aC+bC. \end{align*} Thus the equality is justified for both $K$ and $L$. [/guided] [/step] [step:Apply Brunn-Minkowski to the scaled interpolation sets] Define the sets \begin{align*} A := (1-\lambda)K_s, \qquad B := \lambda K_t. \end{align*} Since $K$ and $L$ are convex bodies, $K_s$ and $K_t$ are compact convex subsets of $\mathbb{R}^n$; nonnegative scalar dilations preserve compactness and convexity, so $A$ and $B$ are compact convex subsets of $\mathbb{R}^n$. By the [Brunn-Minkowski Inequality](/theorems/???), \begin{align*} \mathcal{L}^n(A+B)^{1/n} \geq \mathcal{L}^n(A)^{1/n}+\mathcal{L}^n(B)^{1/n}. \end{align*} Using the identity from the previous step, $A+B=K_r$. For any compact set $E \subset \mathbb{R}^n$ and any scalar $c \geq 0$, \begin{align*} \mathcal{L}^n(cE)^{1/n}=c\,\mathcal{L}^n(E)^{1/n}. \end{align*} For $c>0$ this follows from the linear change of variables $x=cy$, whose Jacobian determinant is $c^n$; for $c=0$, the set $cE$ is a singleton and has $\mathcal{L}^n$-measure $0$. Hence \begin{align*} \mathcal{L}^n(K_r)^{1/n} &= \mathcal{L}^n(A+B)^{1/n} \\ &\geq \mathcal{L}^n((1-\lambda)K_s)^{1/n} +\mathcal{L}^n(\lambda K_t)^{1/n} \\ &= (1-\lambda)\mathcal{L}^n(K_s)^{1/n} +\lambda\mathcal{L}^n(K_t)^{1/n}. \end{align*} [guided] Define \begin{align*} A := (1-\lambda)K_s, \qquad B := \lambda K_t. \end{align*} These are compact convex subsets of $\mathbb{R}^n$, because $K_s$ and $K_t$ are compact and convex, and scalar dilations preserve compactness and convexity. Thus the [Brunn-Minkowski Inequality](/theorems/???) applies to $A$ and $B$, giving \begin{align*} \mathcal{L}^n(A+B)^{1/n} \geq \mathcal{L}^n(A)^{1/n}+\mathcal{L}^n(B)^{1/n}. \end{align*} The previous step identified the Minkowski sum on the left: \begin{align*} A+B=(1-\lambda)K_s+\lambda K_t=K_r. \end{align*} Therefore \begin{align*} \mathcal{L}^n(K_r)^{1/n} \geq \mathcal{L}^n((1-\lambda)K_s)^{1/n} + \mathcal{L}^n(\lambda K_t)^{1/n}. \end{align*} It remains to simplify the two scaled-volume terms. For any compact set $E \subset \mathbb{R}^n$ and scalar $c \geq 0$, \begin{align*} \mathcal{L}^n(cE)^{1/n}=c\,\mathcal{L}^n(E)^{1/n}. \end{align*} When $c>0$, this is the standard scaling rule for Lebesgue measure under the [linear map](/page/Linear%20Map) \begin{align*} S_c: \mathbb{R}^n &\to \mathbb{R}^n \\ y &\mapsto cy. \end{align*} The Jacobian determinant of $S_c$ is $c^n$, so \begin{align*} \mathcal{L}^n(cE)=c^n\mathcal{L}^n(E). \end{align*} Taking $n$th roots gives the displayed homogeneity identity. When $c=0$, the set $cE$ is the singleton $\{0\}$, which has $\mathcal{L}^n$-measure $0$, so the same identity holds. Applying this with $E=K_s$, $c=1-\lambda$, and then with $E=K_t$, $c=\lambda$, we obtain \begin{align*} \mathcal{L}^n(K_r)^{1/n} &\geq \mathcal{L}^n((1-\lambda)K_s)^{1/n} + \mathcal{L}^n(\lambda K_t)^{1/n} \\ &= (1-\lambda)\mathcal{L}^n(K_s)^{1/n} + \lambda\mathcal{L}^n(K_t)^{1/n}. \end{align*} This is exactly the concavity inequality for the volume radius at the parameters $s,t,\lambda$. [/guided] [/step] [step:Conclude concavity and recover the endpoint inequality] Since $s,t,\lambda \in [0,1]$ were arbitrary, and $r=(1-\lambda)s+\lambda t$, the preceding estimate proves \begin{align*} \rho((1-\lambda)s+\lambda t) \geq (1-\lambda)\rho(s)+\lambda\rho(t), \end{align*} so $\rho$ is concave on $[0,1]$. Taking $s=0$ and $t=1$ gives $K_0=K$ and $K_1=L$, hence for every $\lambda \in [0,1]$, \begin{align*} \mathcal{L}^n(K_\lambda)^{1/n} \geq (1-\lambda)\mathcal{L}^n(K)^{1/n} + \lambda\mathcal{L}^n(L)^{1/n}. \end{align*} Renaming $\lambda$ as $t$ gives \begin{align*} \mathcal{L}^n((1-t)K+tL)^{1/n} \geq (1-t)\mathcal{L}^n(K)^{1/n} + t\mathcal{L}^n(L)^{1/n}, \end{align*} which is the stated particular inequality. [/step]