Find all triples $(x,y,z)\in \Bbb{R}$ that satisfy the following conditions: The question is to find all real numbers solutions to the system of equations:  


$y=\Large\frac{4x^2}{4x^2+1}$,
$z=\Large\frac{4y^2}{4y^2+1}$,
$x=\Large\frac{4z^2}{4z^2+1}$,  


This seems simple enough, so I tried substituting the values of x, y and z into the different equations but I only ended up with a huge degree 8 equation which clearly doesn't seem like the right approach. I really have no idea on how to go about solving this if substitution is not the answer.
Any help would be greatly appreciated :)
We can note that $4a^2+1 \ge 4a \ \ \forall a \in R$. Also, since $\frac{4a^2}{4a^2+1} \ge 0$, then we know that $x,y,z \ge 0$

Therefore $y=\frac{4x^2}{4x^2+1} \le \frac{4x^2}{4x}=x$ for nonezero values of $x$. Similarly $z \le y$ and $x \le y$. Therefore $x \le y \le z \le x \implies x=y=z$.

Now solving equation $a=\frac{4a^2}{4a^2+1} \implies 4a^2+1=4a \implies a = \frac{1}{2} \implies x=y=z=\frac{1}{2}$.

Finally, we assumed that numbers are non-zero, so we should include solution $(0,0,0)$
$\bullet\; $ Clearly $x=y=z=0$ are the solution of system of eqn

$\bullet\; $ If $x,y,z\neq 0\; $ Then


$\displaystyle y=\frac{4x^2}{4x^2+1}\Longrightarrow \frac{1}{y}=1+\frac{1}{4x^2}$,
$\displaystyle z=\frac{4y^2}{4y^2+1}\Longrightarrow \frac{1}{z}=1+\frac{1}{4y^2},$
$\displaystyle x=\frac{4z^2}{4z^2+1}\Longrightarrow \frac{1}{x}=1+\frac{1}{4z^2}$,  


Adding all three 

$\displaystyle \bigg(1-\frac{1}{2x}\bigg)^2+\bigg(1-\frac{1}{2y}\bigg)^2+\bigg(1-\frac{1}{2z}\bigg)^2=0$

which is possible when

$\displaystyle 1-\frac{1}{2x}=0,1-\frac{1}{2y}=0,1-\frac{1}{2z}=0$

System of equation have $\displaystyle (x,y,z)=\bigg(\frac{1}{2},\frac{1}{2},\frac{1}{2}\bigg)$ as solution also