Proof: Since $x^{2}+y^{2}+z^{2}=1$, then,
$$\begin{array}{l} g=x+y+z-x y z \\ \geqslant x+y+z-3 x y z \\ \geqslant 3 \sqrt[3]{x y z}-3 x y z \geqslant 0 . \end{array}$$

Thus, $g \geqslant 1$ is equivalent to (homogenization)
$$\begin{array}{l} G(x, y, z) \\ =\left[(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)-x y z\right]^{2}- \\ \quad\left(x^{2}+y^{2}+z^{2}\right)^{3} \\ \geqslant 0 . \end{array}$$

Decomposing $G(x, y, z)$, we get
$$\begin{array}{l} G(x, y, z) \\ =2 g_{6,2}+2 g_{6,3}+8 g_{6,4}+8 g_{6,5}+ \\ \quad 12 g_{6,6}+37 g_{6,7} \end{array}$$
$\geqslant 0$. This problem is difficult if you don't know the minimum value is 1.