Proving $|A+B| \geq |A|+|B| - 1$ for $A,B \subseteq \mathbb{Z}$ finite and non-empty

Question

Let $A,B \subseteq \mathbb{Z}$ be finite and non-empty, and define $A+B = \{a + b : a \in A, b \in B\}$.

Is there a nice way to prove that $|A+B| \geq |A|+|B| - 1$?

Answer 1

Suppose $A, B$ are both non-empty and finite. If we put $A' = \{a - n \mid a \in A\}$ for some $n \in \mathbb{Z}$, then one can easily show that $|A + B| = |A' + B|$. Hence, by translating $A$ and $B$ such that $\max A = \min B = 0$, we see that $A$ and $B$ are both subsets of $A + B$, hence $$ |A + B| \geq |A \cup B| = |A| + |B| - |A \cap B| = |A| + |B| - 1 $$ as $A \cap B = \{0\}$.

Answer 2

I presume $A$ and $B$ are finite and non-empty.

Write $A=\{a_1,\ldots,a_m\}$ where $a_1<\cdots<a_m$ and $B=\{b_1,\ldots,b_n\}$ where $b_1<\cdots<b_n$. Consider $a_1+b_1,a_1+b_2,\ldots,a_1+b_n,a_2+b_n,\ldots,a_m+b_n$.