TITLE: How does the area of $A+A$ compare to the area of $A-A$?
QUESTION [9 upvotes]: The Minkowski sum of two sets $A$ and $B$ in the plane is defined as 
$$A+B = \{ a + b \mid a \in A, b \in B \}.$$
The Minkowski difference $A-B$ is defined similarly.   
For any convex set $A$, is it always true that $$|A-A| \ge |A+A|?$$
For example, if $A$ is a triangle, then $|A - A| = \frac{3}{2} |A + A|$. If $A$ is symmetric about a point, then $|A-A| = |A+A|$.

REPLY [2 votes]: Yes.  By the Brunn–Minkowski inequality, $|A-A|^{1/2}\geq|A|^{1/2}+|-A|^{1/2}$, so $|A-A|\geq 4|A|$.  The hypotheses of this theorem do not include convexity, although they do include compactness.
On the other hand, under the assumption that $A$ is convex, $A+A = 2A$, and $|A+A|=4|A|$. Therefore $|A-A|\geq |A+A|$.