4. 156 Solve the system of equations
$$\left\{\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=y \\
\frac{4 y^{2}}{1+4 y^{2}}=z \\
\frac{4 z^{2}}{1+4 z^{2}}=x
\end{array}\right.$$

for all real solutions, and prove that your solution is correct.
［Solution］ If $x=y$, then
$$\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=x, \\
4 x^{2}=x+4 x^{3}, \\
x\left(4 x^{2}-4 x+1\right)=0, \\
x=0 \text { or } \frac{1}{2},
\end{array}$$

Thus, we get $x=y=z=0$ or $x=y=z=\frac{1}{2}$.
If $x>y$, then
$$\begin{array}{l}
\frac{4 z^{2}}{1+4 z^{2}}>\frac{4 x^{2}}{1+4 x^{2}} \\
1-\frac{1}{1+4 z^{2}}>1-\frac{1}{1+4 x^{2}}
\end{array}$$

Noting that from the original equation, we have $x \geqslant 0, y \geqslant 0, z \geqslant 0$, thus from the above inequality, we get
$$z>x$$

By the cyclic symmetry of the original system of equations, from $z>x$ we can deduce $y>z$. Therefore, we have
$$x>y>z>x$$

Contradiction.
If $x<y$, then similarly we can deduce
$$x<y<z<x$$

Contradiction.
Therefore, the original system of equations has only two solutions:
$$(x, y, z)=(0,0,0) \text { and }(x, y, z)=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) .$$