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?

all of x,y,z have the same sign.
assume x,y,z>0.
assume $x\geq y\geq z$
$$\frac{y}{z}+\frac{z}{y}=2x^2\geq 2y^2=\frac{z}{x}+\frac{x}{z}$$
$$\frac{y}{z}\geq \frac{x}{z}$$
$$y\geq x$$
$$x=y$$
$$2z^2=\frac{x}{y}+\frac{y}{x}=2$$
$$z=1$$
$$2x^2=x+1/x$$
$$2x^3=x^2+1$$
$$2x^3-x^2-1=0$$
$$(x-1)(2x^2+x+1)=0$$
$$x=1$$
$$x=y=z=1$$
dropping the assumption that all are positive, (x,y,z)=(1,1,1) or (-1,-1,-1).
there might be a more elegant solution.
