A friend asked me the following problem:

Is it true that for every $X\subset A\subset \mathbb{Z}$, where $A$ is finite and $X$ is non-empty, that $$\frac{|A+X|}{|X|}\geq \frac{|A+A|}{|A|}?$$

Here the notation $A+B$ denotes the set $\{ a+b\ : a\in A, b\in B\}$. It follows from Rusza's triangle inequality that $$\frac{|A+X|}{|X|}\geq \frac{|A+A|}{|A-X|},$$ but since $|A-X|\geq |A|$, this isn't quite strong enough.

It is not hard to show that the inequality holds in either of the extremal cases where $A+A$ is minimal or maximal - that is when $|A+A|=|A|(|A|+1)/2$ or $|A+A|=2|A|-1$.

Is this statement true in general?

Rewriting the desired inequality as $$\frac{|A+X|}{|A+A|}\geq \frac{|X|}{|A|},$$ we are asking if adding $A$ to $X$ and looking at this as a subset of $A+A$ causes the proportion of elements to increase.

**Edit: Numerical Calculations:**

I did some numerical calculations where I let $A$ run through all possible subsets of $\{1,2,3,\dots,n\}$ and $X$ run through all proper subsets of $A$, and I calculated the ratio $$\frac{|A+X||A|}{|A+A||X|}.$$ The minimum of this ratio over all possible combinations of $A,X$ with $X\subsetneq A\subset\{1,2,3\dots,n\}$ appears in the following table:

$$\begin{array}{cc} \text{value of }n\ \ & \text{minimum}\\ 3 & 4/3\\ 4 & 6/5\\ 5 & 8/7\\ 6 & 10/9\\ 7 & 12/11\\ 8 & 14/13\\ 9 & 16/15\\ 10 & 18/17 \end{array} $$ Numerically the slightly stronger estimate $$\frac{|A+X|}{|A+A|}\geq\left(1+\frac{1}{2|X|+1}\right)\frac{|X|}{|A|}$$ seems to hold.

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