Let $A'$ and $B'$ are respectively subsets of $A$ and $B$ such that if $a' \in A'$ and $a'+b' = x$ such that $b' \in B$ then $b' \in B'$. Then the inequality becomes $|A',B'| |A+B| \le |A-B|^2$. So $|A',B'||A+B|$ counts the number of the set $(a',b',t_{a,b})$, where $t_{a,b} \in A+B$ and $a,b$ are unique pair in $A$ and $B$ respectively satisfying $a+b= t$. Now map this to the set $(a'-b, a-b') \in (A-B)^2$. Assume $(a'' - b_0, a_0 - b'') = (a'-b, a-b')$, then we have $a_0+b_0 = a+b$ implying $a_0 = a ; b_0 = b; a'= a''$ and $b'= b''$, so the map is one-one and we are done.