Feel free to scroll down to the "Question" section if you're familiar with the notation of Tao and Vu's Additive Combinatorics, which I believe is standard notation for the field.
Notation
Let $Z$ be an additive (abelian group). For sets $A,B\subset Z$, we define the sum-sets and difference sets: $$ A+B = \{a+b\in Z; a\in A, b\in B\} $$ and $$ A-B = \{a-b\in Z; a\in A, b\in B\}. $$ We write $2A$ for $A+A$. (Note that this is not dilation, and $A+A$ is in general much larger than $A$). Notice also that $A-B = A+(-B)$, where $-B$ is the reflection of $B$ around $0$.
If $G\subset Z$ is symmetric (so $G=-G$), $0\in G$, and $2G$ may be covered by less than $K$ translates of $G$, then we say that $G$ is a $K$-approximate group. Another way to write this last statement is that there exists a set $X\subset Z$ such that $|X|\le K$ and $$ 2G \subset G+X. $$
Question
The question I have has to do with a part of Exercise 2.4.7 from Tao and Vu's Additive Combinatorics. Suppose $G\subset Z$ is a $K$-approximate group and $H\subset Z$ is a $K'$-approximate group. I have shown that $G+H$ is a $KK'$-approximate group, and using the hint from the textbook I showed that $2G\cap 2H$ is a $(KK')^3$-approximate group. I have also established the cardinality bound $$ \frac{|G||H|}{|G+H|} \leq |2G\cap 2H|. $$ What I am stuck on is proving that $$ |2G\cap 2H| \leq (KK')^3 \frac{|G||H|}{|G+H|}. $$ The hint in the textbook says that I should use the Ruzsa triangle inequality. So my question is:
How can I use the Rusza triangle inequality, along with the fact that $2G\cap 2H$ is a $(KK')^3$-approximate group, to prove the upper bound on the cardinality of $2G\cap 2H$?
My attempt so far I tried to turn the upper bound into something that looked more amenable to using the Rusza triangle inequality, which basically amounts to the statement that for $A,B,C\subset Z$, $$ |A-C| \leq \frac{|A-B||B-C|}{|B|}. $$ The first thing to do is to try to make use of the $(KK')^3$ factor. So I know that $$ |2(2G\cap 2H)| \leq (KK')^3 |2G\cap 2H|. $$ Making this substitution, it would suffice to show that $$ |2G\cap 2H| \leq \frac{|2(2G\cap 2H)||G||H|}{|2G\cap 2H||G+H|}. $$ From here I'm not sure where to go.