3. In the set of real numbers, solve the system

$$
\left\{\begin{array}{l}
x+\sqrt{y}=1 \\
y+\sqrt{z}=1 \\
z+\sqrt{x}=1
\end{array}\right.
$$
Solution. From the condition of the problem, it follows that $x, y, z \geq 0$. Let $x \leq y$. Then $0 \geq x-y=\sqrt{z}-\sqrt{y}$, hence $z \leq y$. From this, $y-z=\sqrt{x}-\sqrt{z} \geq 0$, hence $x \geq z$. From $x-z=\sqrt{x}-\sqrt{y} \leq 0$ it follows that $x \leq z$. Therefore, $x=y=z$. From $x+\sqrt{x}=1$ we get $\sqrt{x}=\frac{-1 \pm \sqrt{5}}{2}$. Since $\sqrt{x} \geq 0, \sqrt{x}=\frac{-1+\sqrt{5}}{2}$, hence $x=y=z=\frac{3-\sqrt{5}}{2}$.

If $x>y$, then similarly to the previous discussion, a contradiction is obtained.

Therefore, the only solutions to the system are $x=y=z=\frac{3-\sqrt{5}}{2}$.