Given two subsets $A$ and $B$ of a finite additive group $Z$, how can one show that there exists an element $x$ in $Z$ such that $$1 - \frac{|A \cap (B + x)|}{|Z|} \le \left(1 - \frac{|A|}{|Z|}\right)\left(1 - \frac{|B|}{|Z|} \right).$$ (This is part of exercise 1.1.4 of Tao and Vu's Additive Combinatorics book).
Any idea how the stated inequality could be proved? The left-hand side seems to be the probability of an element of $Z$ not belonging to $A \cap (B + x)$ and the right-hand side the probability of selecting an element $a$ from $Z \backslash A$ and an element $b$ from $Z \backslash B$.
It seems possible to show that $E(|A \cap (B + x)|)$ among elements $x$ in $Z$ is $|A||B| / |Z|$: for each pair $a$ in $A$ and $b$ in $B$, there is exactly one element $z$ in $Z$ such that $a = b + z$; thus there are $|A||B|$ such triples. Averaging this for the elements of $Z$, we obtain the desired expectation. The first moment method implies an element $x$ in $Z$ such that $$1 - \frac{|A \cap (B + x)|}{|Z|} \le 1 - \frac{|A||B|}{|Z|^2},$$ which seems weaker than the bound stated in the problem.