[proofplan] We prove the two implications separately. If $K$ and $L$ are homothetic, then one is a positive dilation and translation of the other, so the Minkowski sum is the corresponding dilation and translation of a single convex body; translation invariance and homogeneity of Lebesgue measure then give equality. Conversely, equality in Brunn-Minkowski is converted into equality in Minkowski's [first inequality](/theorems/2897) for mixed volumes by applying the mixed-volume identity to $M := K+L$. The equality condition in Minkowski's first inequality then forces both $K$ and $L$ to be homothetic to $M$, hence to each other. [/proofplan]

[step:Record the measure and homothety conventions] For every Lebesgue-measurable set $A \subset \mathbb{R}^n$, write $|A| := \mathcal{L}^n(A)$. A [convex body](/page/Convex%20Body) means a compact convex subset of $\mathbb{R}^n$ with nonempty interior. For subsets $A,B \subset \mathbb{R}^n$, their [Minkowski sum](/page/Minkowski%20Sum) is $A+B := \{a+b : a \in A, b \in B\}$. Since $K$ and $L$ are convex bodies with nonempty interior, they are compact, convex, Lebesgue-measurable sets satisfying \begin{align*} 0 < |K| < \infty, \qquad 0 < |L| < \infty. \end{align*} We use the standard convention that $K$ and $L$ are homothetic if there exist a scalar $a > 0$ and a vector $x_0 \in \mathbb{R}^n$ such that \begin{align*} L = aK + x_0 := \{ak + x_0 : k \in K\}. \end{align*} [/step]

[step:Compute the Minkowski sum when the bodies are homothetic]Assume first that $L = aK + x_0$ for some $a > 0$ and $x_0 \in \mathbb{R}^n$. Then, by the definition of Minkowski addition, \begin{align*} K + L &= K + (aK + x_0) \\ &= \{k_1 + ak_2 + x_0 : k_1,k_2 \in K\}. \end{align*} Since $K$ is convex, the set equality \begin{align*} K + aK = (1+a)K \end{align*} holds: if $k_1,k_2 \in K$, then \begin{align*} \frac{1}{1+a}k_1 + \frac{a}{1+a}k_2 \in K, \end{align*} so $k_1 + ak_2 \in (1+a)K$; the reverse inclusion follows by taking $k_1=k_2=k$ for each $k \in K$. Hence \begin{align*} K+L = (1+a)K + x_0. \end{align*} Translation invariance and dilation homogeneity of $\mathcal{L}^n$ give \begin{align*} |K+L| = |(1+a)K+x_0| = (1+a)^n |K|. \end{align*} Also $|L| = |aK+x_0| = a^n |K|$. Therefore \begin{align*} |K+L|^{1/n} &= (1+a)|K|^{1/n} \\ &= |K|^{1/n} + a|K|^{1/n} \\ &= |K|^{1/n} + |L|^{1/n}. \end{align*} Thus homothetic convex bodies attain equality.[/step]

[guided]Assume that $K$ and $L$ are homothetic. By the convention fixed above, this means there are $a > 0$ and $x_0 \in \mathbb{R}^n$ such that \begin{align*} L = aK + x_0 = \{ak+x_0 : k \in K\}. \end{align*} We now compute the Minkowski sum directly from its definition. Since $K+L = \{u+v : u \in K, v \in L\}$, substituting the displayed description of $L$ gives \begin{align*} K + L &= K + (aK + x_0) \\ &= \{k_1 + ak_2 + x_0 : k_1,k_2 \in K\}. \end{align*} The only geometric input needed here is convexity of $K$. We claim that $K+aK=(1+a)K$. For the inclusion $K+aK \subset (1+a)K$, take $k_1,k_2 \in K$. Because $a>0$, the numbers $1/(1+a)$ and $a/(1+a)$ are nonnegative and sum to $1$; convexity of $K$ gives \begin{align*} \frac{1}{1+a}k_1 + \frac{a}{1+a}k_2 \in K. \end{align*} Multiplying by $1+a$ gives $k_1+ak_2 \in (1+a)K$. For the reverse inclusion, if $k \in K$, then \begin{align*} (1+a)k = k + ak \in K+aK, \end{align*} so $(1+a)K \subset K+aK$. Hence $K+aK=(1+a)K$, and therefore \begin{align*} K+L = (1+a)K + x_0. \end{align*} Lebesgue measure is invariant under translations and scales by the determinant factor under the dilation $x \mapsto (1+a)x$. Since this dilation has determinant $(1+a)^n$, we obtain \begin{align*} |K+L| = |(1+a)K+x_0| = (1+a)^n |K|. \end{align*} Similarly, the translation by $x_0$ does not change measure and the dilation $x \mapsto ax$ has determinant $a^n$, so \begin{align*} |L| = |aK+x_0| = a^n |K|. \end{align*} Taking $n$-th roots, which is valid because $|K|>0$, gives \begin{align*} |K+L|^{1/n} &= (1+a)|K|^{1/n} \\ &= |K|^{1/n} + a|K|^{1/n} \\ &= |K|^{1/n} + |L|^{1/n}. \end{align*} This proves the equality direction for homothetic convex bodies.[/guided]

[step:Reduce equality to equality in Minkowski's first inequality for mixed volumes]Assume conversely that equality holds: \begin{align*} |K+L|^{1/n} = |K|^{1/n}+|L|^{1/n}. \end{align*} Define the convex body $M \subset \mathbb{R}^n$ by $M := K+L$. Let $V(A_1,\dots,A_n)$ denote the mixed volume of convex bodies $A_1,\dots,A_n \subset \mathbb{R}^n$, normalized so that $V(A,\dots,A)=|A|$. By multilinearity of mixed volume with respect to Minkowski addition, \begin{align*} |M| &= V(M,\dots,M) \\ &= V(M,\dots,M,K) + V(M,\dots,M,L), \end{align*} where $M$ occurs $n-1$ times in each term on the last line. Minkowski's first inequality for mixed volumes gives \begin{align*} V(M,\dots,M,K) &\ge |M|^{(n-1)/n}|K|^{1/n}, \\ V(M,\dots,M,L) &\ge |M|^{(n-1)/n}|L|^{1/n}. \end{align*} Adding these inequalities and using the assumed equality gives \begin{align*} |M| &\ge |M|^{(n-1)/n}\bigl(|K|^{1/n}+|L|^{1/n}\bigr) \\ &= |M|^{(n-1)/n}|M|^{1/n} \\ &= |M|. \end{align*} Since the first and last quantities are equal, both applications of Minkowski's first inequality are equalities. The equality condition in Minkowski's first inequality for convex bodies with nonempty interior therefore implies that $K$ is homothetic to $M$ and that $L$ is homothetic to $M$. Homothety is an [equivalence relation](/page/Equivalence%20Relation) among convex bodies with nonempty interior, so $K$ and $L$ are homothetic.[/step]

[guided]Now suppose equality holds: \begin{align*} |K+L|^{1/n} = |K|^{1/n}+|L|^{1/n}. \end{align*} We must avoid citing the equality case of Brunn-Minkowski itself. Instead, we use a distinct equality theorem: the equality condition in Minkowski's first inequality for mixed volumes. Define $M := K+L$. Since the Minkowski sum of compact convex sets is compact and convex, and since $K$ and $L$ have nonempty interior, $M$ is again a convex body with nonempty interior. Let $V(A_1,\dots,A_n)$ denote the mixed volume of convex bodies $A_1,\dots,A_n \subset \mathbb{R}^n$, normalized by $V(A,\dots,A)=|A|$. Mixed volume is multilinear in each argument with respect to Minkowski addition. Applying this multilinearity in the last argument to $M=K+L$ gives \begin{align*} |M| &= V(M,\dots,M) \\ &= V(M,\dots,M,K+L) \\ &= V(M,\dots,M,K) + V(M,\dots,M,L), \end{align*} where $M$ occurs $n-1$ times in every mixed-volume term on the last two lines. Minkowski's first inequality applies because $M$, $K$, and $L$ are convex bodies with nonempty interior, hence all have positive finite Lebesgue measure. It gives \begin{align*} V(M,\dots,M,K) &\ge |M|^{(n-1)/n}|K|^{1/n}, \\ V(M,\dots,M,L) &\ge |M|^{(n-1)/n}|L|^{1/n}. \end{align*} Adding these two inequalities and using the mixed-volume identity above yields \begin{align*} |M| &= V(M,\dots,M,K) + V(M,\dots,M,L) \\ &\ge |M|^{(n-1)/n}\bigl(|K|^{1/n}+|L|^{1/n}\bigr). \end{align*} The assumed Brunn-Minkowski equality says $|K|^{1/n}+|L|^{1/n}=|K+L|^{1/n}=|M|^{1/n}$. Substitution gives \begin{align*} |M| \ge |M|^{(n-1)/n}|M|^{1/n}=|M|. \end{align*} Thus equality must have occurred in both separate applications of Minkowski's first inequality; otherwise their sum would be strictly larger than $|M|$. The equality condition in Minkowski's first inequality now applies. Equality in the pair $(M,K)$ implies that $K$ is homothetic to $M$, and equality in the pair $(M,L)$ implies that $L$ is homothetic to $M$. Since homothety is transitive after composing the corresponding positive dilations and translations, $K$ and $L$ are homothetic. Combining this converse implication with the direct computation for homothetic bodies proves the equivalence claimed in the theorem.[/guided]