Find the minimum value of P

We can make the system equations of the condition more simply by creating 2 new non-negative variables $a$ and $b$ such that $a=\sqrt{x^2+1}$ and $b= \sqrt{y^2-1}$ Then, we have $\left\{\begin{matrix} a+x=y-b & \\ a(1-x)=y(b+1) & \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -x+y=a+b & \\ ax+(b+1)y=a & \end{matrix}\right.$ To solve this system, we can do with different ways. And here is a solution using Cramer's rule $\left\{\begin{matrix} D= \begin{vmatrix} a & b+1\\ -1 & 1 \end{vmatrix}=a+b+1& \\ D_x=\begin{vmatrix} b+1 &a \\ 1 & a+b \end{vmatrix}=b(a+b+1) & \\D_y=\begin{vmatrix} a & a\\ -1 & a+b \end{vmatrix}=a(a+b+1) & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=\dfrac{D_x}{D}=b & \\ y=\dfrac{D_y}{D}=a & \end{matrix}\right.\Rightarrow x^2=y^2-1\left ( \ast \right )$ Thus, we can replace $x^2$ by $y^2-1$ in expression $P$ .

In particular, $P = \dfrac{1}{y^2-1} + y^2 -1$ Moreover, according to $AM-GM$ inequality, we have $P \geq 2$ when $y^2-1 = 1$ or $y=\sqrt{2}$ , thereby determining $x = -1$ .

In other words, the minimum of $P$ is 2 and it occurs when $x=-1, y=\sqrt{2}$ .