Example 19 (2001 National High School Competition Question) The range of the function $y=x+\sqrt{x^{2}-3 x+2}$ is $\qquad$ .
Solve: Fill in $\left[1, \frac{3}{2}\right] \cup[2,+\infty)$. Reason: From $x^{2}-3 x+2 \geqslant 0$, we get $x \leqslant 1$ or $x \geqslant 2$. Also, $x^{2}-3 x+2=\left(x-\frac{3}{2}\right)^{2}-\frac{1}{4}$. Let $x-\frac{3}{2}=\frac{1}{2} \sec \theta, \theta \in\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right)$, then $y=\frac{3}{2}+\frac{1}{2} \sec \theta+\frac{1}{2}|\tan \theta|$.
When $\theta \in\left[0, \frac{\pi}{2}\right]$, $y=\frac{3}{2}+\frac{1}{2} \sec \theta+\frac{1}{2} \tan \theta=\frac{3}{2}+\frac{1}{2} \cdot \frac{1+\sin \theta}{\cos \theta}$
$$
=\frac{3}{2}+\frac{1}{2} \cdot \frac{1-\cos \left(\frac{\pi}{2}+\theta\right)}{\sin \left(\frac{\pi}{2}+\theta\right)}=\frac{3}{2}+\frac{1}{2} \tan \left(\frac{\theta}{2}+\frac{\pi}{4}\right) .
$$

Also, $\frac{\pi}{4} \leqslant \frac{\theta}{2}+\frac{\pi}{4}<\frac{\pi}{2}$, then $\tan \left(\frac{\theta}{2}+\frac{\pi}{4}\right) \geqslant 1, y \geqslant 2$.
Similarly, when $\theta \in\left(\frac{\pi}{2}, \pi\right]$, $1 \leqslant y<\frac{3}{2}$.
Therefore, the range of the function is $\left[1, \frac{3}{2}\right) \cup[2,+\infty)$.
$$
\begin{array}{l}
=\sqrt{108} \cdot \sqrt{2+\sin 2 \beta}-5 \\
\leqslant \sqrt{108} \cdot \sqrt{3}-5=13 .
\end{array}
$$

The above inequality holds with equality if and only if $\sin 2 \beta=1$ and $\sin \alpha=\frac{\sqrt{3}}{3}, \cos \alpha=\frac{\sqrt{6}}{3}$, i.e., $x=3, y=1, z=4$.
Thus, the original system of equations has only one solution: $x=3, y=1, z=4$.