Question: Let $x, y, z \in(0,+\infty)$ and $x^{2}+y^{2}+$ $z^{2}=1$, find the range of the function $f=x+y+z-x y z$.

Find the range of the function $f=x+y+z-x y z$ given that $x, y, z \in(0,+\infty)$ and $x^{2}+y^{2}+z^{2}=1$.
Given $x^{2}+y^{2}+z^{2}=1$, so $y^{2}+z^{2}=1-x^{2}$. Therefore, we can set $y=\sqrt{1-x^{2}} \cos \theta, z=\sqrt{1-x^{2}} \sin \theta, \theta \in\left(0, \frac{\pi}{2}\right)$, thus $f=x+\sqrt{1-x^{2}}(\sin \theta+\cos \theta) - x(1-x^{2}) \sin \theta \cos \theta$.

Let $t=\sin \theta+\cos \theta$, then $\sin \theta \cos \theta=\frac{t^{2}-1}{2}$. Knowing that $\theta \in\left(0, \frac{\pi}{2}\right)$, we have $t=\sin \theta+\cos \theta=\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right) \in(1, \sqrt{2}]$.

Therefore, $f=-\frac{x(1-x^{2})}{2} t^{2}+\sqrt{1-x^{2}} t+x+\frac{x(1-x^{2})}{2}$.

This is a quadratic function in terms of $t$, with a negative coefficient for the quadratic term, and its axis of symmetry is:
$$t=\frac{1}{x \sqrt{1-x^{2}}} \geqslant \frac{1}{\frac{x^{2}+1-x^{2}}{2}}=2$$

Thus, $f$ is monotonically increasing on $(1, \sqrt{2}]$.
So when $t \in(1, \sqrt{2})$, $f>\lim _{t \rightarrow 1} f=x+\sqrt{1-x^{2}}$.

When $t=\sqrt{2}$, $f=\frac{x^{3}}{2}+\frac{x}{2}+\sqrt{2(1-x^{2})}$.
Therefore, $x+\sqrt{1-x^{2}}>1$, so $f>1$.

Also, when $x=y \rightarrow 0, z \rightarrow 1$, $f \rightarrow 1$, so 1 is the optimal lower bound of $f$.