Prove Convex Hull of Minkowski sum I want to prove that the following holds, where the $+$ means Minkowski sum:
$$
conv(A+B)=conv(A)+conv(B)
$$
Let the convex hull of $A+B$ be 
$$
conv(A+B)=\sum_{j,k}\lambda_j\mu_k(a_j+b_k)
$$
I don't know how to continue from here. 
 A: A set $C$ is convex iff for all $t\in (0,1)$ we have $tC+(1-t)C= C$.
It follows that the Minkowski sum of convex sets $A, B$ is convex: 
$$t(A+B)+(1-t)(A+B) = (tA+(1-t)A) + (tB+(1-t)B) = A+B$$
Therefore, for general sets $A,B$ the sum $\operatorname{conv}(A)+\operatorname{conv}(B)$ is convex; and since it contains $A+B$, it also contains the convex hull of $A+B$. One inclusion proved. 
For the opposite  inclusion, pick a point in the convex hull of $A+B$. It is  a convex combination of some points of $A+B$, i.e., $\sum \lambda_k (a_k+b_k)$. Since
$$ \sum \lambda_k (a_k+b_k)= \sum \lambda_k  a_k + \sum \lambda_k  b_k \in \operatorname{conv}(A)+\operatorname{conv}(B)$$
A: $\newcommand{\co}{\operatorname{co}}$ As already noted we have $\co (A+B) \subset \co A +\co B$. Now let $v+w \in \co A + \co B$. Write $v = \sum_i \alpha_i a_i$ and $w = \sum_j \beta_j b_j$. First we have
$$
 v + b_j = \sum_i \alpha_i (a_i + b_j)
$$ so $v + b_j \in \co ( A +B)$. Now we have
$$
v+w = \sum_j \beta_j ( v+ b_j).
$$ So $v+w \in \co( \co (A+B)) = \co (A+B)$ and we are done.
