4. determine all real solutions $(x, y, z)$ of the system

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

## Solution
Obviously, the triples $(x, y, z)=(0,0,0)$ and $(x, y, z)=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ are solutions. We show that they are the only ones. Set $f(t)=4 t^{2} /\left(1+4 t^{2}\right)$. It holds $f(t) \geq 0$ for all $t \in \mathbb{R}$, so $x, y, z$ are non-negative. Furthermore, the following applies

$$
f(t) \leq t, \quad \text { for all } t \geq 0 \text {. }
$$

This is equivalent to $4 t^{2} \leq t\left(1+4 t^{2}\right)$ (note that the denominator of $f(t)$ is positive), and this in turn to $t(2 t-1)^{2} \geq 0$. The latter is correct for $t \geq 0$ and equality only applies for $t=0$ and $t=\frac{1}{2}$. Now assume that there is a solution to the system such that $x \neq 0, \frac{1}{2}$. According to the above considerations

$$
x=f(z) \leq z=f(y) \leq y=f(x)<x
$$

whereby strict inequality applies in the last estimate. This is a contradiction. So $x=0$ or $x=\frac{1}{2}$ and by inserting you can easily calculate that this leads to the two solutions above.