Prove that a set $A$ is convex $(x+y)A = xA + yA$

Question

Consider convex sets $A,B \subset \mathbb{R}^n$ and $x,y \in \mathbb{R}$.

Let $A+B$ be defined as the Minkowski sum of the two sets, that is, $$A+B = \{a+b : a\in A, b\in B\}$$

Let $cA$ be defined as $$cA = \{ca : a \in A\}$$

Prove that $A$ is convex if and only if $(x+y)A = xA + yA$ for every $x,y \geq 0$.

I would usually show an attempt for questions that I ask here but unfortunately, I can't even figure out where to start with this problem. It is difficult to see how can I use the property of convexity to prove the iff statement.

Any tips or hints?

Answer 1

For $a\in A$, then $(x+y)a=xa+xy\in xA+yA$, so $(x+y)A\subseteq xA+yA$.

For $a,b\in A$, since $A$ is convex, then $\dfrac{x}{x+y}a+\dfrac{y}{x+y}b\in A$, so $xa+yb\in(x+y)A$ and hence $xA+yA\subseteq(x+y)A$.