2. Find all real solutions to the system of equations
$$
\left\{\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=y, \\
\frac{4 y^{2}}{1+4 y^{2}}=z, \\
\frac{4 z^{2}}{1+4 z^{2}}=x .
\end{array}\right.
$$

and prove that your solution is correct.
2. Solution 1 Let $x=y$, then $x=y=z$. From $4 x^{2}=x+4 x^{3}$, we get $x=0$ or $\frac{1}{2}$.

Let $x>y$, then $1-\frac{1}{1+4 z^{2}}>1-\frac{1}{1+4 x^{2}}$, thus $z>x$, and hence $1-\frac{1}{1+4 y^{2}}>1-\frac{1}{1+4 z^{2}}$, which implies $y>z$. But $x>y>z>x$ is impossible.
When $x>0, y>0, z>0$, (1) + (2) + (3) gives $x+y+z=\frac{4 x^{2}}{1+4 x^{2}}+\frac{4 y^{2}}{1+4 y^{2}}+\frac{4 z^{2}}{1+4 z^{2}} \leqslant \frac{4 x^{2}}{4 x}+\frac{4 y^{2}}{4 y}+$ $\frac{4 z^{2}}{4 z}=x+y+z$.
The equality holds if and only if $x=y=z=\frac{1}{2}$.
Upon verification, $x=\frac{1}{2}, y=\frac{1}{2}, z=\frac{1}{2}$ is a solution to the system of equations.
Therefore, the system of equations has only two solutions.