Algebra vs. Trigonometry

Sorry, something was wrong with my solution so I had to repost the whole thing

First, $x-\sqrt{y-1}(z+1)=\sqrt{yz^2+2yz-2z^2-4z}\iff x=\sqrt{yz^2+2yz-2z^2-4z}+\sqrt{y-1}(z+1)$

$=\sqrt{(2-y)(-z^2-2z)}+\sqrt{y-1}(z+1)$

$=\sqrt{2-y}.\sqrt{-z^2-2z}+\sqrt{y-1}(z+1)$

$=\sqrt{1-(y-1)}.\sqrt{1-(z+1)^2}+\sqrt{y-1}(z+1)$

$1\le y\le 2 \implies 0\le \sqrt{y-1} \le 1$ . Let $\sqrt{y-1}=\sin B;\sqrt{1-(y-1)}=\cos B$ with $B \in [0;90\circ]$ .

By the same way, let $z+1=\cos C;\sqrt{1-(z+1)^2}=\sin C$ .

$\implies x=\sin B\cos C+\cos B\sin C=\sin(B+C)$

$1\le y\le 2 \implies P=x^{2}+\sqrt{7y}-z^2-2z$

$=x^{2}+2\sqrt{\frac{7}{4}y}-z^2-2z \le x^2+y+\frac{7}{4}-z^2-2z$

(the equality holds when $y=\frac{7}{4}$ )

$=x^2+[1-(z+1)^2]+y-1+\frac{11}{4}$

$=\sin^2(B+C)+\sin^2B+\sin^2C+\frac{11}{4}$

$=\sin^2(B+C)+1-\dfrac{1}{2}(\cos2B+\cos2C)+\frac{11}{4}$

$=2-\cos^2(B+C)-\cos(B+C)\cos(B-C)+\frac{11}{4}$

$=5-(\cos(B+C)+\frac{1}{2}\cos(B-C))^2-\frac{1}{4}\sin^2(B-C) \le 5$

The equality holds when

$\begin{aligned} \begin{cases} \sin^2(B-C) = 0 \\ \cos(B+C)+\dfrac{1}{2}\cos(B-C) = 0 \end{cases} \end{aligned}$ $\iff \begin{aligned} \begin{cases} x = \frac{\sqrt{3}}{2} \\ y = \frac{7}{4} \\ z= -\frac{1}{2} \end{cases} \end{aligned}$

To sum up, the maximum of $P$ is $\boxed{5}$ .