Let $a$ and $b$ be positive reals. Prove $$\frac{1}{3} > \frac{2ab^2}{(a + \sqrt{a^2 + b^2})^3}\;.$$
I am a high school student preparing for math contests, and this question was sent to me by one of my classmates on a contest math training course, but I wasn't able to solve it. I do not know where he got this problem from. I tried using AM-GM, since $2ab \leq 2a^2 + b^2$, but I couldn't progress further. Also, I also tried to manipulate the expression to use the HM-AM-GM-QM inequality, and I got to this:
$\sqrt{\dfrac{a^2 + b^2}{2}} > \left(\dfrac{b\sqrt{3}}{a + \sqrt{a^2 + b^2}} - 1\right)\left(\dfrac{b\sqrt{3}}{a + \sqrt{a^2 + b^2}} + \dfrac{a\sqrt{2}}{2}\right)$
Which didn't help much (and I am not even sure if this is correct). After these attempts, I got stuck. If anyone could give a hint, it would be very appreciated. Full solutions are also fine, but I would like to try to work out with the hints first.