Let $A$ be a finite set of real numbers. Is it always the case that $|AA+A| \geq |A/A|$?

In the line above, $AA+A:=\{ab+c:a,b,c \in A \}$, while $A/A:=\{a/b:a,b \in A, b\neq 0 \}$ is the ratio set.

This is closely related to a previous question about the relative sizes of the sets $AA+A$ and $A+A$. See Is the set $ AA+A $ always at least as large as $ A+A $?.


1 Answer 1


Just in case anyone out there was wondering, it turns out that the answer to this question is "no".

Earlier this year, a paper of myself, Imre Rusza, Chun-Yen Shen and Ilya Shkredov was posted. One of the results in the paper was a construction of a set of reals $A$ with $|AA+A|=o(|A|^2)$. For this particular example, one has $|A/A| = \Omega(|A|^2)$.

I suspect that the following weaker conjecture holds: for all $\epsilon >0$ there is an absolute constant $c=c(\epsilon)$ with $|AA+A| \geq c|A/A|^{1-\epsilon}$.


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