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 $&gt;0$.

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 $x^2&gt;yz$ we get $$x^2y^2z^2&gt;xyxzzy$$
a contradiction. So $x^2=yz$ and similary for other, so we have equality sign every where and thus $$1+4x^2 = 4x\implies x={1\over 2} = y=z$$
Taking the resultant of equations 1 and 2 with respect to $z$, we find 
$$ 64\,{x}^{2}{y}^{5}+32\,{x}^{2}{y}^{3}+16\,{y}^{5}-64\,{x}^{4}+4\,{x}^{
2}y+8\,{y}^{3}+y
=0 $$
Taking the resultant of equations 1 and 3 with respect to $z$, we find
$$ 4 x^3 y + x y - 4 y^2 + x = 0$$
Taking the resultant of these two with respect to $y$, we find
$x = 0$ or $x = 1/2$ or 
$$ 65536\,{x}^{18}+65536\,{x}^{17}+147456\,{x}^{16}+131072\,{x}^{15}+
156672\,{x}^{14}+205824\,{x}^{13}+221440\,{x}^{12}+251904\,{x}^{11}+
234624\,{x}^{10}+203392\,{x}^{9}+179232\,{x}^{8}+159104\,{x}^{7}+
131524\,{x}^{6}+39460\,{x}^{5}+10681\,{x}^{4}+2800\,{x}^{3}+680\,{x}^{
2}+128\,x+16
 = 0$$
This last polynomial has no real roots. The real solutions are $(0,0,0)$ corresponding to $x=0$ and $(1/2,1/2,1/2)$ corresponding to $x=1/2$.