Solve for positive real solutions of cyclic equations $x+y^2+2xy=9$, $y+z^2+2yz=47$, $z+x^2+2xz=16$ 
Solve over positive reals $$x+y^2+2xy=9$$
$$y+z^2+2yz=47$$
$$z+x^2+2xz=16$$

With standard manipulation we get that $x+y+z=8$. Thus $x=8-y-z$ and we have two equations,
$$(8-y-z)^2+y^2+2(8-y-z)y=9$$
$$y+z^2+2yz=47$$
Which becomes
$$-y^2-2yz+15y-z=1$$
$$y+z^2+2yz=47$$
Adding up gets
$$-y^2+16y+z^2-z=1$$
$$\implies z^2=z+1+y^2-16y$$
And this substitutes to 
$$y+z+1+y^2-16y+2yz=47$$
$$\implies y+\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)+1+y^2-16y+2y\bigg(\frac{-y^2+15y-1}{2y+1}\bigg)=47$$
However, this results into a nasty quartic which I don't know how to simplify. Thanks!

$\begin{cases}
x+y^2+2xy=9\\
y+z^2+2yz=47\\
z+x^2+2xz=16
\end{cases} \implies
\begin{cases}
(2 x + 2 y)^2 - (2 x - 1)^2 = 35\\
(2 y + 2 z)^2 - (2 y - 1)^2 = 187\\
(2 z + 2 x)^2 - (2 z - 1)^2 = 63
\end{cases} \implies
\begin{cases}
x=1\\
y=2\\
z=5
\end{cases}$
