If $X, Y$ are subsets of an abelian group, we denote $X - Y = \left\{ x - y \ \lvert \ x \in X, \ y \in Y \right\}$.
Question: Let $A, B, C$ be non-empty subsets of an abelian group $G$. Is the inequality $$\lvert A \rvert \lvert B - C \rvert \leq \lvert A - B \rvert \lvert A - C \rvert$$ true without assuming the Axiom of Choice?
The Ruzsa triangle inequality states that for all finite subsets $A, B, C$ of some abelian group, the inequality $$\lvert A \rvert \lvert B - C \rvert \leq \lvert A - B \rvert \lvert A - C \rvert$$ holds. The proof, as given in the linked Wikipedia article, is as follows: for each element $x \in B - C$, arbitrarily choose $b(x) \in B, \ c(x) \in C$ such that $b(x) - c(x) = x$. Now, the function $$f : A \times \left( B - C \right) \to \left( A - B \right) \times \left( A - C \right)$$ which is defined by $$f(a, x) = \left( a - b(x), a - c(x) \right)$$ is injective, as $\left( a - c(x) \right) - \left( a - b(x) \right) = b(x) - c(x)$, and so we can recover $b(x) - c(x) = x$, which lets us recover $b(x)$, which lets us recover $a$.
If we try to do this for the case where $A, B, C$ are infinite, then when trying to choose $b(x), c(x)$ we need to use the Axiom of Choice, and so I wondered if this is possible to do without Choice?
By the way, assuming the axiom of Choice the infinite case has an easy proof, because if either $B$ or $C$ are infinite, then $$\max \left( \lvert B \rvert, \lvert C \rvert \right) = \lvert B \times C \rvert \geq \lvert B - C \rvert \geq \max \left( \lvert B \rvert, \lvert C \rvert \right)$$ and from here it's easy to finish, but of course this doesn't work well without Choice.