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. 



A sumset inequality
The proposed inequality is not true. I do not claim originality for this example: all I have done is take Seva's observation that the inequality would imply an improvement $|3A|\leq K^2|A|$ given $|2A|\leq K|A|$, find a standard counterexample to this hypothetical improvement, and then simplify it a bit for the present situation.


I will work inside in $\mathbf{Z}^2$, but this makes no difference because one can always just look at the image in $\mathbf{Z}$ under $(x,y)\mapsto x + My$ for large $M$.


Let $X = [x]^2$ and
$$A = X \cup \{(j,0): j\in [Kx]\}\cup\{(0,j): j\in [Kx]\}.$$
Then for $x$ large enough $|A+X|/|X|\approx 2K$ and $|A+A|/|A|\approx K^2$.


One can get something from Petridis's results though, namely that if $X\subset A$ and $|A+X|\leq K|X|$ then $|A+A|\leq K^2|A|$.


