Finding all possible complex solutions to the system $xy = \sqrt 2(x + y)$, $yz = \sqrt 3(y + z)$, $zx = \sqrt 5(z + x)$ 
Find all possible complex solutions to:
$$\begin{cases}xy = \sqrt 2(x + y) \\
yz = \sqrt 3(y + z)\\
zx = \sqrt 5(z + x)\end{cases}$$

I have some ideas:
$$xy-\sqrt 2y=\sqrt 2x\implies y=\frac{\sqrt 2x}{x-\sqrt 2} $$
and
$$yz-\sqrt 3y=\sqrt 3z\implies y=\frac{\sqrt 3z}{z-\sqrt 3} $$
and
$$xz-\sqrt 5x=\sqrt 5z\implies x=\frac{\sqrt 5z}{z-\sqrt 5} $$
So I got
$$y=\frac{\sqrt 2x}{x-\sqrt 2}=\frac{\sqrt 3z}{z-\sqrt 3}=\frac{\sqrt 2×\frac{\sqrt 5z}{z-\sqrt 5}}{\frac{\sqrt 5z}{z-\sqrt 5}-\sqrt 2}$$
From here we can go to a solution, but that brings up really scary equations.  Is there any clever way to solve this?  That is a minor Olympiad problem.

Observe that,
If $x=0$ or $y=0$ or $z=0$, you will get $x=y=z=0$. So, we can assume $x,y,z\neq 0$.
$$\begin{cases}\frac 1x+\frac 1y=\frac {\sqrt 2}{2}\\
\frac 1y+\frac 1z=\frac {\sqrt 3}{3}\\ 
\frac 1x+\frac 1z=\frac {\sqrt 5}{5}\end{cases}\implies \frac 1x+\frac 1y+\frac 1z=\frac {\frac {\sqrt 2}{2}+\frac {\sqrt 3}{3}+\frac {\sqrt 5}{5}}{2}=a$$
Finally, we have
$$\frac 1x=a-\frac {\sqrt 3}{3}\implies x=\frac{1}{a-\frac {\sqrt 3}{3}}$$
Note that,  $y$ and $z$ can be found in the same way.
