Double estimates relating Ruzsa distance and doubling constant I am trying to solve the following exercise (2.3.16) from Tao-Vu book.

Let $A,B$ be additive sets with common ambient group $Z$. Show that
$\sigma[A\cup B]\leq e^{d(A,B)}+2e^{4d(A,B)}$. In the converse direction, establish
the estimate $$d(A,B)\leq \log \sigma[A\cup B]+\dfrac{1}{2}\log
 \dfrac{|A\cup B|}{|A|}+\dfrac{1}{2}\log \dfrac{|A\cup B|}{|B|}.$$

My approach: i) We note that the first inequality can be written as follows: $$\dfrac{|A\cup B+A\cup B|}{|A\cup B|}\leq \dfrac{|A-B|}{|A|^{1/2}|B|^{1/2}}+2\dfrac{|A-B|^4}{|A|^2|B|^2}.$$ It is easy to show that $|A\cup B+A\cup B|\leq |A+A|+|B+B|+|A+B|.$ By Ruzsa triangle inequality we obtain: $d(A,-A)\leq d(A,-B)+d(-B,-A) \Leftrightarrow |A+A|\leq \dfrac{|A+B||A-B|}{|B|}$. In the same way one can show that $|B+B|\leq \dfrac{|A+B||A-B|}{|A|}.$ Since $d(A,-B)\leq 3d(A,B)$ (see Corollary 2.12) then $|A+B|\leq \dfrac{|A-B|^3}{|A||B|}.$ Hence we obtain $$|A+A|\leq \dfrac{|A-B|^4}{|A||B|^2} \ \ \text{and} \ \ |B+B|\leq \dfrac{|A-B|^4}{|A|^2|B|}.$$
Therefore, $$\dfrac{|A\cup B+A\cup B|}{|A\cup B|}\leq 2\dfrac{|A-B|^4}{|A|^2|B|^2}+\dfrac{|A+B|}{|A\cup B|},$$ since $|A|,|B|\leq |A\cup B|.$ However, I cannot prove that $\dfrac{|A+B|}{|A\cup B|}\leq \dfrac{|A-B|}{|A|^{1/2}|B|^{1/2}}$ and I firmly believe that this bound is not true at all because if we take $A=B$ then it is equivalent to $\sigma[A]\leq \delta[A],$ where $\sigma[A]=\dfrac{|2A|}{|A|}$ and $\delta[A]=\dfrac{|A-A|}{|A|}$ are doubling and difference constants.
ii) The second inequality is equivalent to $|A-B|\leq |A\cup B+A\cup B|$ and I have no idea how to prove that. I was trying to prove using Ruzsa triangle inequality but I failed.
It would be great to see how to solve this exercise.
 A: The inequality in the "converse direction" has a typo since, for example, if $A=B$ it says $\delta[A]\leq \sigma[A]$, which need not be true. The inequality should instead involve $c\log\sigma[A\cup B]$, where $c$ is some absolute constant. I leave the identification of this constant as an exercise (use Exercise 2.3.4).
As for the first inequality, you are of course correct that one cannot have $|A+B|/|A\cup B|\leq e^{d(A,B)}$ in general. I don't see offhand how to improve your argument to effectively absorb more data into the $2e^{4d(A,B)}$ term; nor do I have a counterexample to show that it cannot be done. On the other hand, the format of the proposed inequality is very suggestive of your strategy, in which one shows that $|A+A|/|A\cup B|$ and $|B+B|/|A\cup B|$ are both bounded by $e^{4d(A,B)}$, while $|A+B|/|A\cup B|$ is bounded by some other term. So I wonder if there isn't also a typo here, and instead of $e^{d(A,B)}$, it should be something that actually does bound $|A+B|/|A\cup B|$ (for example, $e^{3d(A,B)}$). This would fit with the spirit of the exercise, which is just to show that if $A$ and $B$ are close in Rusza distance, then $A\cup B$ has small doubling. So to summarize, what I am saying is that the following inequality is true (essentially by what you've already written), and suffices to establish the purpose of the exercise:
$$
\sigma[A\cup B]\leq e^{3d(A,B)}+2e^{4d(A,B)}.
$$
