Solve in $\mathbb{R^{3}}$ : $\begin{cases}(1+4x^{2})y=4z^{2}\\(1+4y^{2})z=4x^{2}\\(1+4z^{2})x=4y^{2}\end{cases}$ Solve the system in $\mathbb{R^{3}}$ : 

$$\begin{cases}(1+4x^{2})y=4z^{2}\\(1+4y^{2})z=4x^{2}\\(1+4z^{2})x=4y^{2}\end{cases}$$

My try : 
By imaging I see $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ is a solution! 
From a first equation : 
$$x=\frac{1}{2}\sqrt{\frac{4z^{2}}{y}-1}$$
So by second equation : 
$$y=\frac{1}{2}\sqrt{\frac{\frac{4z^2}{y}-1}{z}-1}$$
But after applied I get difficult equation for $z$ ?

We see that all are nonnegative. 
If one of them is $0$ then all are $x=y=z=0$. Now suppose all are $>0$.
Now $1+4x^2\geq 4x$ so $$4z^2=y(1+4x^2)\geq 4xy\implies z^2\geq xy$$
In the same manner we get $y^2\geq xz$ and $x^2\geq yz$ so, if say $x^2>yz$ we get $$x^2y^2z^2>xyxzzy$$
a contradiction. So $x^2=yz$ and similary for other, so we have equality sign every where and thus $$1+4x^2 = 4x\implies x={1\over 2} = y=z$$
