RHS is increasing (derivative calculation), ez. (Everyone at the competition who solved this problem has solved it like this, try to guess author's solution.)

Here are the details. To show $x\sqrt{\frac{x^3+2}{2(x^3+1)}} > y\sqrt{\frac{y^3+2}{2(y^3+1)}}$ for $x>y>0$ either use derivative on $f(x) = x\sqrt{\frac{x^3+2}{2(x^3+1)}}$ or note that it is equivalent to $(x-y)(x^4 - x^2y^2 + y^4 + x^4y^3 + x^3y^4 + x^3y + xy^3 + 2x + 2y) > 0$ which is true since $x^4 - x^2y^2 + y^4 = (x^2 - y^2)^2 + x^2y^2 > 0$ and everything else is positive. Now we show the main statement by induction on $n$. The case $n=1$ is clear. Now if we assume $a_n > \sqrt{\frac{3}{n}}$ for some $n$, then the initial claim gives $$ a_{n+1} = a_n\sqrt{\frac{a_n^3+2}{2(a_n^3+1)}} = \sqrt{\frac{a_n^2}{2}\left(1 + \frac{1}{a_n^3+1}\right)} > \sqrt{\frac{3}{2n}\left(1 + \frac{1}{1 + \frac{3\sqrt{3}}{n\sqrt{n}}}\right)} $$and now we want the latter to be at least $\sqrt{\frac{3}{n+1}}$. This is equivalent to $(2z+\sqrt{3})(z - \sqrt{3})^2 \geq 0$ where $z = \sqrt{n}$, done!

I think, the most interesting question about this problem is what exactly is the rate of convergence of $a_n$. It converges to $0$ as $n\to \infty$. Indeed, $\frac{a_{n}^3+2}{2(a_{n}^3+1)}<1,$ thus $a_n$ decreases and no other limit is possible. One may think that $a_n=O(\frac{1}{\sqrt n})$ but but it's not the case. In fact, the estimate given is not sharp, the sequence decreases far more slowly. It could have been more interesting if it were to prove $\displaystyle a_n\ge C\frac{1}{\sqrt[3] n}$ for an absolute constant $C>0.$ One may look at the details here.

Assassino9931 wrote: Now if we assume $a_n > \sqrt{\frac{3}{n}}$ for some $n$, then the initial claim gives $$ a_{n+1} = a_n\sqrt{\frac{a_n^3+2}{2(a_n^3+1)}} = \sqrt{\frac{a_n^2}{2}\left(1 + \frac{1}{a_n^3+1}\right)} > \sqrt{\frac{3}{2n}\left(1 + \frac{1}{1 + \frac{3\sqrt{3}}{n\sqrt{n}}}\right)} $$ Are you sure that simply plugging in the bound on $a_n$ is correct?

@above Absolutely, since $f(a_n) > f\left(\sqrt{\frac{3}{n}}\right)$ whenever $a_n > \sqrt{\frac{3}{n}}$, as I have justified that $f$ is an increasing function.

My bad, I somehow managed to ignore the part before that completely.