(1) Since $x^{2}+y^{2}+z^{2}=1$, we have $x+y+z \geq 3 \sqrt[3]{x y z} \geq 3 x y z$, so $g=x+y+z-x y z>0$, and $\sigma_{1}^{2}-2 \sigma_{2}=x^{2}+y^{2}+z^{2}=1$, which gives $g=\sigma_{1}\left(\sigma_{1}^{2}-2 \sigma_{2}\right)-\sigma_{3}$,
$$\begin{aligned} g \geq 1 & \Longleftrightarrow\left(\sigma_{1}^{3}-2 \sigma_{1} \sigma_{2}-\sigma_{3}\right)^{2} \geq\left(\sigma_{1}^{2}-2 \sigma_{2}\right)^{3} \\ & \Longleftrightarrow 2 \sigma_{1}^{4} \sigma_{2}-8 \sigma_{1}^{2} \sigma_{2}^{2}+8 \sigma_{2}^{3}-2 \sigma_{1} \sigma_{3}\left(\sigma_{1}^{2}-2 \sigma_{2}\right)+\sigma_{3}^{2} \geq 0 \\ & \Longleftrightarrow 2 f_{0,2}^{(6)}+2 f_{0,3}^{(6)}+8 f_{0,4}^{(6)}+8 f_{1,1}^{(6)}+12 f_{1,2}^{(6)}+37 \sigma_{3}^{2} \geq 0 \end{aligned}$$

It is clear that (3.2.16) holds. Therefore, (3.2.15) holds, and we know that when $y=z=0, x=1$, $g$ attains its minimum value of 1.
(2) Using the same technique,
$$\begin{aligned} g \leq \frac{8 \sqrt{3}}{9} & \Longleftrightarrow 81\left(\sigma_{1}^{3}-2 \sigma_{1} \sigma_{2}-\sigma_{3}\right)^{2} \leq 192\left(\sigma_{1}^{2}-2 \sigma_{2}\right)^{3} \\ & \Longleftrightarrow L_{9}=37 f_{0,1}^{(6)}-17 f_{0,2}^{(6)}+94 f_{0,3}^{(6)}+80 f_{0,4}^{(6)}+117 f_{1,1}^{(6)}-28 f_{1,2}^{(6)} \geq 0 \end{aligned}$$

Since
$$\begin{array}{l} L_{9} \geq 17\left(f_{0,1}^{(6)}+f_{1,1}^{(6)}-f_{0,2}^{(6)}\right)+28\left(f_{0,4}^{(6)}+f_{1,1}^{(6)}-f_{1,2}^{(6)}\right) \\ f_{0.1}^{(6)}+f_{1,1}^{(6)}-f_{0.2}^{(6)}=\sum x^{2}(x-y)^{2}(x-z)^{2} \geq 0 \\ f_{0,4}^{(6)}+f_{1,1}^{(6)}-f_{1,2}^{(6)} \geq 2 \sqrt{f_{0.4}^{(6)} f_{1,1}^{(6)}}-f_{1,2}^{(6)} \geq 0 \end{array}$$

The correctness of (3.2.18) is guaranteed by (3.1.8), so $L_{9} \geq 0, g \leq \frac{8 \sqrt{3}}{9}$ holds. Q.E.D.