For any two finite subsets $A,B$ of an abelian group, does the following ineqality hold?
$$|A+B|^2 |A-B|^2 \geq |A+A||A-A||B+B||B-B| \ $$
I’m interested in finding out if there are sumset inequalities that are sharper than the triangle inequalities and that show that some quantity related to sums and differences is smaller when the sums and differences are over the same sets.
Smaller examples don’t seem to work: $|A-B|^2 \geq |A-A||B-B|$ fails when $B = -A$ and $|A+A| < |A-A|$.
$|A+B|^2 \geq |A+A||B+B|$ doesn’t work when $B = -A$ and $|A-A| < |A+A|$.
The candidate given in the title combines the two and avoids these difficulties because of its symmetry. It also holds when $A$ is an arithmetic progression or a subspace (if such finite subspaces exist) and B is a random set.