Let $F$ be a finite field of odd size $q$, and $\phi_0 : F \mapsto F$ be any map from $F$ to itself. For each $a \in F$, set $\phi_a : x \in F \mapsto \phi_0 (x) + ax $.

When $\phi_0 : x \mapsto x^2 $ , each image $\phi_a (F)$ has size $\frac{q+1}{2}$. It turns out that for any $\phi_0$, there's always some $a \in F$ such that $| \phi_a(F) | \geq \frac{q+1}{2}$.

But the only proof I know is somewhat artificial : it relies on the observation that $ K =\bigcup_{a \in F} \phi_a(F)^n $ is a Kakeya set of dimension $n$ over the finite field $F$. By subsequent improvements of Dvir's work on such sets, it is known that $K$ must have $\geq \left( \frac{q^2}{2q-1} \right)^n $ elements. As $n$ is arbitrary and $\frac{q^2}{2q-1} > \frac{q+1}{2} -1 $, this yields the above claim.

I've found no direct proof so far. It seems MO might be the right place to ask for such a proof.