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.