A sumset inequality 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: This is a very incomplete answer at this stage, but it establishes some connections worth recording. 
A well-known lemma by Petridis says that if $A$ and $B$ are finite, non-empty subsets of an abelian group such that $|X+B|/|X|\ge|A+B|/|A|$ holds for every non-empty subset $X\subset A$, then for every finite set $C$ one has $|A+B+C||A|\le|A+B||A+C|$. Your inequality (if true) says that the assumption of Petridis' lemma always holds true in the situation where $B=A$ and the underlying group is the group of integers. As a result, 
  $$ |2A+C||A|\le|2A||A+C| $$
for any finite integer sets $A$ and $C$. Letting here $C=(n-2)A$ with integer $n\ge 3$, we conclude that if $|2A|=K|A|$, then $|nA|\le K^{n-1}|A|$. The "standard" estimate here is $K^n|A|$, and I doubt it can be improved to $K^{n-1}|A|$ - though, maybe, the integer case is special in this respect.
A: 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 $|A+A|\leq K^2|A|$.
