[proofplan] We compare the one-variable volume polynomial $t \mapsto V(K+tL)$ with the Brunn--Minkowski lower bound for its $n$th root. Differentiating at $t=0$ identifies the right derivative with the mixed volume $V(K[n-1],L)$, and the derivative comparison gives the inequality after raising to the $n$th power. For the equality case, the same one-variable function is concave by Brunn--Minkowski; equality in the derivative forces it to coincide with its Brunn--Minkowski supporting line, so the equality case of Brunn--Minkowski implies homothety. [/proofplan]

[step:Handle the one-dimensional case and reduce to positive volume] If $n=1$, then the mixed-volume notation $V(K[n-1],L)$ means $V(L)$, since there are no copies of $K$ and one copy of $L$. Hence \begin{align*} V(K[n-1],L)^n &=V(L), \\ V(K)^{n-1}V(L) &=V(K)^0V(L)=V(L), \end{align*} so the inequality holds with equality. If, in addition, $K$ and $L$ have non-empty interior, then they are non-degenerate compact intervals in $\mathbb{R}$. Writing $K=[a,b]$ and $L=[c,d]$ with $a<b$ and $c<d$, define \begin{align*} \lambda:=\frac{d-c}{b-a}>0. \end{align*} Then $L=(c-\lambda a)+\lambda K$, so $K$ and $L$ are homothetic. Thus the theorem is proved when $n=1$. Assume for the rest of the proof that $n\geq 2$. If $V(K)=0$, then $n-1>0$ and the asserted inequality becomes \begin{align*} V(K[n-1],L)^n \geq 0, \end{align*} which follows from non-negativity of mixed volumes. Assume now that $V(K)>0$. Since $K$ is a convex body in $\mathbb{R}^n$, this means that $K$ has non-empty interior. Define the Minkowski-sum path \begin{align*} A: [0,\infty) &\to \{\text{convex bodies in } \mathbb{R}^n\} \\ t &\mapsto K+tL, \end{align*} where \begin{align*} K+tL := \{x+ty \in \mathbb{R}^n : x \in K,\ y \in L\}. \end{align*} Also define \begin{align*} f: [0,\infty) &\to [0,\infty) \\ t &\mapsto V(K+tL)^{1/n}. \end{align*} Because $K$ has non-empty interior, $V(K)>0$, and hence $f(0)=V(K)^{1/n}>0$. [/step]

[step:Differentiate the Brunn Minkowski lower bound at $t=0$]By the Brunn--[Minkowski inequality](/theorems/517) for convex bodies (citing a result not yet in the wiki: Brunn--Minkowski Theorem), for every $t \geq 0$ we have \begin{align*} V(K+tL)^{1/n} \geq V(K)^{1/n} + V(tL)^{1/n}. \end{align*} Since $V(tL)=t^nV(L)$ by homogeneity of Lebesgue measure under the dilation map $y \mapsto ty$, this becomes \begin{align*} f(t) \geq f(0)+tV(L)^{1/n}. \end{align*} Therefore, for every $t>0$, \begin{align*} \frac{f(t)-f(0)}{t} \geq V(L)^{1/n}. \end{align*} Taking the limit inferior as $t \downarrow 0$ gives \begin{align*} f'_+(0) \geq V(L)^{1/n}, \end{align*} provided the right derivative exists. The mixed-volume polynomial expansion gives \begin{align*} V(K+tL)=\sum_{j=0}^{n} \binom{n}{j} V(K[n-j],L[j])t^j, \end{align*} so the right derivative of $t \mapsto V(K+tL)$ at $0$ exists and equals \begin{align*} \frac{d}{dt}\Big|_{t=0+}V(K+tL)=nV(K[n-1],L). \end{align*} Since $r \mapsto r^{1/n}$ is differentiable at $r=V(K)>0$, the chain rule gives \begin{align*} f'_+(0) &= \frac{1}{n}V(K)^{1/n-1}\cdot nV(K[n-1],L) \\ &= V(K)^{1/n-1}V(K[n-1],L). \end{align*} Combining the two displays yields \begin{align*} V(K)^{1/n-1}V(K[n-1],L) \geq V(L)^{1/n}. \end{align*}[/step]

[guided]The purpose of this step is to convert the global Brunn--Minkowski inequality into an infinitesimal statement at $K$. We look only along the ray of convex bodies $K+tL$ because the coefficient of $t$ in its volume polynomial is exactly the mixed volume $nV(K[n-1],L)$. By the Brunn--Minkowski inequality for convex bodies (citing a result not yet in the wiki: Brunn--Minkowski Theorem), applied to the convex bodies $K$ and $tL$, we have for every $t \geq 0$: \begin{align*} V(K+tL)^{1/n} \geq V(K)^{1/n}+V(tL)^{1/n}. \end{align*} The dilation map \begin{align*} D_t: \mathbb{R}^n &\to \mathbb{R}^n \\ y &\mapsto ty \end{align*} has Jacobian determinant $t^n$, so $\mathcal{L}^n(tL)=t^n\mathcal{L}^n(L)$. Therefore \begin{align*} V(tL)^{1/n}=tV(L)^{1/n}. \end{align*} Thus, with \begin{align*} f: [0,\infty) &\to [0,\infty) \\ t &\mapsto V(K+tL)^{1/n}, \end{align*} we obtain \begin{align*} f(t) \geq f(0)+tV(L)^{1/n}. \end{align*} Subtracting $f(0)$ and dividing by $t>0$ gives \begin{align*} \frac{f(t)-f(0)}{t} \geq V(L)^{1/n}. \end{align*} We now compute the right derivative of $f$ at $0$. The mixed-volume polynomial expansion says \begin{align*} V(K+tL)=\sum_{j=0}^{n}\binom{n}{j}V(K[n-j],L[j])t^j. \end{align*} The coefficient of $t$ is $nV(K[n-1],L)$, so \begin{align*} \frac{d}{dt}\Big|_{t=0+}V(K+tL)=nV(K[n-1],L). \end{align*} Because $V(K)>0$, the scalar function $r \mapsto r^{1/n}$ is differentiable at $r=V(K)$. Applying the ordinary one-variable chain rule to $f(t)=V(K+tL)^{1/n}$ gives \begin{align*} f'_+(0) &= \frac{1}{n}V(K)^{1/n-1}\cdot nV(K[n-1],L) \\ &= V(K)^{1/n-1}V(K[n-1],L). \end{align*} Taking the limit in the difference-quotient inequality therefore yields \begin{align*} V(K)^{1/n-1}V(K[n-1],L) \geq V(L)^{1/n}. \end{align*}[/guided]

[step:Rearrange the derivative inequality into Minkowski first inequality] Since $V(K)>0$, multiply \begin{align*} V(K)^{1/n-1}V(K[n-1],L) \geq V(L)^{1/n} \end{align*} by $V(K)^{1-1/n}$. This gives \begin{align*} V(K[n-1],L) \geq V(K)^{(n-1)/n}V(L)^{1/n}. \end{align*} Both sides are non-negative, so raising to the $n$th power preserves the inequality: \begin{align*} V(K[n-1],L)^n \geq V(K)^{n-1}V(L). \end{align*} Together with the one-dimensional and zero-volume cases handled above, this proves the inequality for all convex bodies $K,L \subset \mathbb{R}^n$. [/step]

[step:Verify equality when the bodies are homothetic] Assume $K$ and $L$ have non-empty interior and are homothetic. Then there exist a point $a \in \mathbb{R}^n$ and a scalar $\lambda>0$ such that \begin{align*} L=a+\lambda K. \end{align*} Mixed volume is invariant under translating any argument and is homogeneous of degree one in each argument. Hence \begin{align*} V(K[n-1],L)=V(K[n-1],a+\lambda K)=\lambda V(K[n]). \end{align*} Since $V(K[n])=V(K)$, we have \begin{align*} V(K[n-1],L)=\lambda V(K). \end{align*} Also $V(L)=V(a+\lambda K)=\lambda^nV(K)$. Therefore \begin{align*} V(K[n-1],L)^n &=(\lambda V(K))^n \\ &=\lambda^nV(K)^n \\ &=V(K)^{n-1}V(L). \end{align*} Thus equality holds for homothetic convex bodies with non-empty interior. [/step]

[step:Force Brunn Minkowski equality from equality in the derivative comparison]Assume now that $K$ and $L$ have non-empty interior and equality holds: \begin{align*} V(K[n-1],L)^n=V(K)^{n-1}V(L). \end{align*} Since $V(K)>0$ and $V(L)>0$, the derivative inequality from the previous steps is an equality: \begin{align*} f'_+(0)=V(L)^{1/n}. \end{align*} The Brunn--Minkowski inequality also implies that $f$ is concave on $[0,\infty)$. Indeed, for $s,t \geq 0$ and $\theta \in [0,1]$, \begin{align*} K+((1-\theta)s+\theta t)L =(1-\theta)(K+sL)+\theta(K+tL), \end{align*} and Brunn--Minkowski gives \begin{align*} f((1-\theta)s+\theta t) \geq (1-\theta)f(s)+\theta f(t). \end{align*} For a concave function on $[0,\infty)$, the right derivative at $0$ gives an upper supporting line: \begin{align*} f(t)\leq f(0)+tf'_+(0) \end{align*} for every $t\geq 0$. Hence \begin{align*} f(t)\leq V(K)^{1/n}+tV(L)^{1/n}. \end{align*} The Brunn--Minkowski lower bound gives the reverse inequality, so for every $t>0$, \begin{align*} V(K+tL)^{1/n}=V(K)^{1/n}+tV(L)^{1/n}. \end{align*} Thus equality holds in Brunn--Minkowski for the pair $K$ and $tL$ for every $t>0$. By the equality case in the Brunn--Minkowski theorem for full-dimensional convex bodies (citing a result not yet in the wiki: Equality Case in Brunn--Minkowski), $K$ and $tL$ are homothetic. Since dilation by $t>0$ preserves homothety, $K$ and $L$ are homothetic.[/step]

[guided]We now prove the converse equality statement. The assumption is \begin{align*} V(K[n-1],L)^n=V(K)^{n-1}V(L). \end{align*} Because both $K$ and $L$ have non-empty interior, $V(K)>0$ and $V(L)>0$. Therefore the inequality \begin{align*} V(K)^{1/n-1}V(K[n-1],L) \geq V(L)^{1/n} \end{align*} obtained from differentiating Brunn--Minkowski must actually be an equality: \begin{align*} f'_+(0)=V(L)^{1/n}. \end{align*} The key point is that equality of the derivative is stronger than it first appears because $f$ is concave. We verify concavity directly from Brunn--Minkowski. Let $s,t \geq 0$ and $\theta \in [0,1]$. The Minkowski combination of the two bodies $K+sL$ and $K+tL$ is \begin{align*} (1-\theta)(K+sL)+\theta(K+tL) =K+((1-\theta)s+\theta t)L. \end{align*} Applying Brunn--Minkowski to these two convex bodies gives \begin{align*} f((1-\theta)s+\theta t) \geq (1-\theta)f(s)+\theta f(t). \end{align*} This is exactly concavity of $f$ on $[0,\infty)$. For a concave function, the slope of a secant line starting at $0$ is bounded above by the right derivative at $0$. Therefore, for every $t>0$, \begin{align*} \frac{f(t)-f(0)}{t}\leq f'_+(0). \end{align*} Using $f'_+(0)=V(L)^{1/n}$, we get \begin{align*} f(t)\leq f(0)+tV(L)^{1/n} =V(K)^{1/n}+tV(L)^{1/n}. \end{align*} But Brunn--Minkowski already gave the opposite inequality: \begin{align*} f(t)\geq V(K)^{1/n}+tV(L)^{1/n}. \end{align*} Hence both inequalities are equalities: \begin{align*} V(K+tL)^{1/n}=V(K)^{1/n}+tV(L)^{1/n} \end{align*} for every $t>0$. This is equality in Brunn--Minkowski for the pair $K$ and $tL$. Since $K$ and $L$ have non-empty interior, so do $K$ and $tL$. The equality case in the Brunn--Minkowski theorem for full-dimensional convex bodies (citing a result not yet in the wiki: Equality Case in Brunn--Minkowski) therefore implies that $K$ and $tL$ are homothetic. Multiplying a convex body by the positive scalar $t$ does not change whether it is homothetic to $K$, so $K$ and $L$ are homothetic.[/guided]