What convex function should be used here?

Let,
$f(x) = \sqrt{\left(\dfrac{x^{3}}{x^{3}+1}\right)} , x \ge 0$

It can be proved that $f(x)$ is concave for $x \ge 0$ According to Jensen's inequality,
$f\left(\dfrac{1}{a}\right) +f\left(\dfrac{1}{b}\right) + f\left(\dfrac{1}{c}\right) \le 3f\left(\dfrac{\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}}{3}\right)$
$\therefore \displaystyle \sum_{cyclic} \dfrac{1}{\sqrt{a^{3}+1}} \le 3f(1)$
$\therefore \displaystyle \sum_{cyclic}\dfrac{1}{\sqrt{a^{3} + 1}} \le \dfrac{3}{\sqrt{2}}$

$x + y = 2 + 3 = 5$

By AM-GM, we can obtain the values of $a$ , $b$ , and $c$ which are $a=b=c=1$ . Using those values, $\frac{x}{\sqrt{y}}=\frac{3}{\sqrt{2}}$ and $x+y=\boxed{5}$ .

Feel free to fix my solutions as it is slightly incomplete :)