More than a system

The constrain of $x, y$ are $x\geq 1$ and $y\geq -1$ . First, we set $a=\sqrt{x-1}\geq 0$ and $b=\sqrt{y+1}\geq 0$ , we'll get $\begin{cases} a+2b=b^2+\frac{5}{4} \ (1) \\ \frac{a^2}{1+b}+\frac{b^2}{1+a}=\frac{1}{3} \ (2)\end{cases}$ Now we consider $(2)$ , we see that $LHS_{(2)}\stackrel{Titu's}\geq\frac{(a+b)^2}{2+a+b} \ (3)$ From $(1)$ we get $a+b=b^2-b+\frac{5}{4}$ , so now we set $t=a+b=b^2-b+\frac{5}{4}$ $RHS_{(3)}=\frac{t^2}{t+2}=t+2+\frac{4}{t+2}-4$ We have $t=b^2-b+\frac{5}{4}\geq 1$ , so it can be predicted that $RHS_{(3)}$ reaches its minimum value when $t=1$ , so we do a little manipulation, combining with AM-GM $RHS_{(3)}=t+2+\frac{9}{t+2}-\frac{5}{t+2}-4\geq 2\sqrt{(t+2)\frac{9}{t+2}}-\frac{5}{1+2}-4=\frac{1}{3}=RHS_{(2)}$ So finally we get $LHS_{(2)}\geq RHS_{(2)}$ , the equality happens when $\begin{cases} a=b \\ t=b^2-b+\frac{5}{4}=1\end{cases}$ Solving the system and we'll get $(x,y)=\big(\frac{5}{4},\frac{-3}{4}\big)$ as the solution, therefore the product is $-0.9375$