How does strong convexity behave under Minkowski sums? Suppose we have two convex sets $A,B$. It is straightforward to show the sum $A+B = \{a+b:a \in A,b\in B\}$ is convex. For take any $c=a+b$ and $c'=a'+b'$ and consider the convex combination
$$xc+(1-x)c' = x(a+b)+(1-x)(a'+b')$$ $$= \big(xa+(1-x)a'\big) + \big( xb+(1-x)b'\big).$$
By convexity the two summand are in $A,B$ respectively. Hence the sum is in $A+B$ as required.
Now suppose $A$ is $\alpha$-strongly convex and  $B$ is $\beta$-strongly convex. That means the for each $a,a' \in A$ and $x \in [0,1]$ that the ball centred at $xa+(1-x)a'$ of radius $\alpha x(1-x)$ is contained in $A$. Likewise for each $b,b' \in B$ and $x \in [0,1]$ that the ball centred at $xb+(1-x)b'$ of radius $\beta x(1-x)$ is contained in $B$.
Can we say anything about the strong-convexity parameter of $A+B$? By that I mean $\inf\{ \gamma \ge 0: A+B$ is $\gamma$-strongly convex $\}$.
 A: I'm going for the largest strong convexity constant, since any smaller value also works.
If $A$ is $\alpha$-strongly convex and $B$ is $\beta$-strongly convex, then $A+B$ is $\gamma:=\max\{\alpha,\beta\}$-strongly convex.
Proof: Let $a_1,a_2\in A$, let $b_1,b_2\in B$, and let $\eta\in[0,1]$. For notational simplicity, I will set $a:=\eta a_1+(1-\eta)a_2$ and $b:=\eta b_1+(1-\eta)b_2$. Now suppose that $z$ is in the ball centered at $\eta(a_1+b_1) + (1-\eta)(a_2+b_2)=a+b$ with radius $\eta(1-\eta)\gamma$. It suffices to show that $z\in A+B$. By the construction of $\gamma$ and definition of $z$, either
\begin{equation}
\|a+b-z\|=\|a - \left(z - b\right)\| \leq \eta(1-\eta)\gamma=\eta(1-\eta)\max\{\alpha,\beta\} =\eta(1-\eta)\alpha, \tag{1}
\end{equation}
or,
\begin{equation}
\|a+b-z\|=\|b - \left(z - a\right)\|\leq \eta(1-\eta)\beta. \tag{2}
\end{equation}
If (1) holds, then $z-b$ is in the ball centered at $a$ with the proper radius. Since $A$ is strongly convex, this implies $z-b\in A$ and hence $z\in A+b\subset A+B$. Likewise, if (2) holds, then $z-a$ is in the ball centered at $b$ with the proper radius. Since $B$ is strongly convex, we find $z-a\in B$ and hence $z\in B+a\subset A+B$.
