An algebra problem by P C

First, we see that $y\geq 0; 0\leq x\leq 2$ . Set $a=\sqrt{x}, b=\sqrt{y}, c=\sqrt{2-x}$ and we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ , the expression is rewritten as $\sum_{cyc}\frac{a}{a^4+1+2ab}=\sum_{cyc}\frac{1}{a^3+\frac{1}{a}+2b}$ By AM-GM we get $\sum_{cyc}\frac{1}{a^3+\frac{1}{a}+2b}\leq \frac{1}{2}\sum_{cyc}\frac{1}{a+b}$ Based on this inequality $\frac{1}{t}+\frac{1}{u}\geq\frac{4}{t+u}$ we'll have $\frac{1}{2}\sum_{cyc}\frac{1}{a+b}\leq \frac{1}{8}.2\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)=\frac{3}{4}=0.75$ The equality holds when $x=y=1$