Sumset Bound |A+B| ≥ |A|+|B|-1 in Additive Combinatorics

Recall that $A+B$ denotes the set of all real numbers $a+b$ for $a$ in $A$ and $b$ in $B$.

Let $|A|=m$, and $|B|=n$ where $m$ and $n$ are positive integers and $A$, $B$ are sets of reals. Say $A={a_1, a_2,...,a_m}$ and $B={b_1, b_2,...,b_n}$ where the sets are in increasing order. Then note that

$$a_1+b_1<a_1+b_2<a_1+b_3< ... <a_1+b_{n-1} < a_1+b_n,$$ $$a_1+b_n<a_2+b_n<a_3+b_n< ... <a_{m-1}+b_n < a_m+b_n.$$

There are $m+n-1$ terms here, and they are all different. So $|A+B| \ge |A|+|B|-1$ in the reals, and certainly in any subset of the reals.

This is also true modulo a prime $p$ (with the technicality that if $|A|+|B| \ge p+1$, then $|A+B| = p$), and it is called the Cauchy-Davenport Theorem.

(2009 USAMO #2) Let $n$ be a positive integer. Determine the size of the largest subset of $\{-n, -n+1, \ldots , n-1, n\}$ which does not contain three elements $a, b, c$ (not necessarily distinct) satisfying $a+b+c=0$.