<s>[INST] 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 :)
 [/INST] $\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
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