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 $$1+4x^2 = 4x\implies x={1\over 2} = y=z$$
Hint
What if $x=0?$
For nontrivial cases $x,y,z>0$
$$(1+4x^2)(1+4y^2)(1+4z^2)=?$$
$$(1-2x)^2\ge0\implies1+4x^2\ge4x$$
Check when the equality can occur
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 $x=1/2$.
