Solving the system $2x^2 = \frac{y}{z}+\frac{z}{y}$, $2y^2 = \frac{z}{x}+\frac{x}{z}$, $2z^2 = \frac{x}{y}+\frac{y}{x}$ 
Find all triplets $\{ x, y, z\}$, such that all three of them are real and nonzero, and satisfies:
$$2x^2 = \frac{y}{z}+\frac{z}{y}$$
$$2y^2 = \frac{z}{x}+\frac{x}{z}$$
$$2z^2 = \frac{x}{y}+\frac{y}{x}$$

I'm stuck in this problem. My first thought was that the only solutions that are possible are $\{ 1, 1, 1 \}$ and $\{ -1, -1, -1\}$, but I do not know how to prove that these are the only solutions. I know for a fact that there seems something "fishy" about the fractions, and I am thinking of changing those into:
$$2x^2yz=y^2+z^2$$
$$2y^2xz=x^2+z^2$$
$$2z^2xy=y^2+x^2$$
but I do not know how to proceed from here. Can anybody give me a hint on how to proceed?


$$2x^2yz=y^2+z^2 \\ 2y^2xz=x^2+z^2$$

Subtracting the two:
$$
\require{cancel}
2xyz(x-y) = y^2 + \bcancel{z^2} - x^2 - \bcancel{z^2} = (y-x)(y+x)
$$
If $\,y \ne x\,$, after canceling out the $\,(x-y)\,$ factor $\,2xyz=-x-y\,$, then it follows by symmetry that $\,2xyz =-y-z=-z-x\,$, so $\,x = y\,$ i.e. contradiction.
This leaves the case $\,x=y\,$, so $\,x=y=z\,$ by symmetry, and the equations reduce to $\,2x^2=2\,$.
