$\newcommand{\R}{\mathbb{R}}$Actually, for compact subsets $A, B\subset \R^n$, the inequality \begin{equation*} m(A+B)^{1/n} \ge m(A)^{1/n} + m(B)^{1/n} \end{equation*} is ultimately nothing but the convexity of $\log(1+e^x)$ in disguise (whose convexity in turn follows immediately from the AM-GM inequality).

This idea, and the entire inductive proof may be found in Theorem 2.3.9, Notions of Convexity, L. Hörmander (2007, Birkhäuser); this is a superb book which anybody with interest in convexity should definitely own!

The idea of the proof is to show this inequality when $A$ and $B$ are unions of finitely many disjoint products of intervals, by induction over the number of intervals that make up $A$ and $B$.

I'll just cite the base case: here $A$ and $B$ are intervals of side lengths $a_1,\ldots,a_k$ and $b_1,\ldots,b_k$, then

\begin{equation*} m(A)^{1/n} +m(B)^{1/n} = \prod\nolimits_j a_j^{1/n} + \prod\nolimits_j b_j^{1/n} \ge \prod\nolimits_j (a_j+b_j)^{1/n}, \end{equation*} where the final inequality follows from convexity of $\log(1+e^x)$.

$\newcommand{\R}{\mathbb{R}}$Actually, for compact subsets $A, B\subset \R^n$, the inequality \begin{equation*} m(A+B)^{1/n} \ge m(A)^{1/n} + m(B)^{1/n} \end{equation*} is ultimately nothing but the convexity of $\log(1+e^x)$ in disguise (whose convexity in turn follows immediately from the AM-GM inequality).

This idea, and the entire inductive proof may be found in Theorem 2.3.9, Notions of Convexity, L. Hörmander (2007, Birkhäuser); this is a superb book which anybody with interest in convexity should definitely own!

The idea of the proof is to show this inequality when $A$ and $B$ are unions of finitely many disjoint products of intervals, by induction over the number of intervals that make up $A$ and $B$.

I'll just cite the base case: here $A$ and $B$ are intervals of side lengths $a_1,\ldots,a_k$ and $b_1,\ldots,b_k$, then

\begin{equation*} m(A)^{1/n} +m(B)^{1/n} = \prod\nolimits_j a_j^{1/n} + \prod\nolimits_j b_j^{1/n} \le \prod\nolimits_j (a_j+b_j)^{1/n}, \end{equation*} where the final inequality follows from homogeneity of terms on both sides of the inequality upon combination with convexity of $\log(1+e^x)$.

$\newcommand{\R}{\mathbb{R}}$Actually, for compact subsets $A, B\subset \R^n$, the inequality \begin{equation*} m(A+B)^{1/n} \ge m(A)^{1/n} + m(B)^{1/n} \end{equation*} is ultimately nothing but the convexity of $\log(1+e^x)$ in disguise (whose convexity in turn follows immediately from the AM-GM inequality).

This idea, and the entire inductive proof may found in Theorem 2.3.9, Notions of Convexity, L. Hörmander (2007, Birkhäuser); this is a superb book which anybody with interest in convexity should definitely own!

The idea of the proof is to show this inequality when $A$ and $B$ are unions of finitely many disjoint products of intervals, by induction over the number of intervals that make up $A$ and $B$.

I'll just cite the base case: here $A$ and $B$ are intervals of side lengths $a_1,\ldots,a_k$ and $b_1,\ldots,b_k$, then

\begin{equation*} m(A)^{1/n} +m(B)^{1/n} = \prod\nolimits_j a_j^{1/n} + \prod\nolimits_j b_j^{1/n} \ge \prod\nolimits_j (a_j+b_j)^{1/n}, \end{equation*} where the final inequality follows from convexity of $\log(1+e^x)$.

$\newcommand{\R}{\mathbb{R}}$Actually, for compact subsets $A, B\subset \R^n$, the inequality \begin{equation*} m(A+B)^{1/n} \ge m(A)^{1/n} + m(B)^{1/n} \end{equation*} is ultimately nothing but the convexity of $\log(1+e^x)$ in disguise (whose convexity in turn follows immediately from the AM-GM inequality).

This idea, and the entire inductive proof may be found in Theorem 2.3.9, Notions of Convexity, L. Hörmander (2007, Birkhäuser); this is a superb book which anybody with interest in convexity should definitely own!

The idea of the proof is to show this inequality when $A$ and $B$ are unions of finitely many disjoint products of intervals, by induction over the number of intervals that make up $A$ and $B$.

I'll just cite the base case: here $A$ and $B$ are intervals of side lengths $a_1,\ldots,a_k$ and $b_1,\ldots,b_k$, then

\begin{equation*} m(A)^{1/n} +m(B)^{1/n} = \prod\nolimits_j a_j^{1/n} + \prod\nolimits_j b_j^{1/n} \ge \prod\nolimits_j (a_j+b_j)^{1/n}, \end{equation*} where the final inequality follows from convexity of $\log(1+e^x)$.