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\begin{document}

\title{An inequality due to Ruzsa}
\author{Dakota Blair}
\maketitle

\begin{thm*}
Let $G$ be a group and $X, A_1, A_2$ be finite subsets of $G$.
\[
|X| |A_1 + A_2| \leq |A_1 - X| |X + A_2|
\]
\end{thm*}

\begin{proof}
Let $A = A_1 + A_2$.
Pick $r : A \to A_1 \times A_2$
such that $r(a) = (a_{1,a}, a_{2,a})$ and $a = a_{1,a} + a_{2,a}$.
Define a map 
\begin{align*}
f: X \times A &\to (A_1 - X) \times (X + A_2)\\
(x, a) = (x, a_{1,a} + a_{2,a}) & \mapsto (a_{1,a} - x, x + a_{2,a}).
\end{align*}
This map is injective.
Let $(x, a), (x', a') \in X \times A$ and $f((x, a)) = f((x',a'))$.
Then
\[
f((x, a)) = (a_{1,a} - x, x + a_{2,a}) = (a'_{1,a'} - x', x' + a'_{2,a'})
= f((x',a')).
\]
Thus
\begin{align*}
\tag{1} a_{1,a} - x & = a'_{1,a'} - x'\\
\tag{2} x + a_{2,a} & = x' + a'_{2,a'}.
\end{align*}
so $(1) + (2)$ shows $a = a_{1,a} + a_{2,a} = a'_{1,a'} + a'_{2,a'} = a'$.
Since $r$ is injective,
$r(a) = (a_{1,a}, a_{2,a}) = (a'_{1,a'}, a'_{2,a'}) = r(a')$.
In particular $a_{2,a} = a'_{2,a'}$, so $(2)$ implies $x = x'$.
Therefore $(x, a) = (x',a')$. Hence $f$ is injective and
\[
|X| |A_1 + A_2| \leq |A_1 - X| |X + A_2|.
\]
\end{proof}
\end{document}
