Example: Let $a, b, c \in \mathbf{R}^{+}, \alpha \in \mathbf{R}$, and assume
$$
\begin{array}{c}
f(\alpha)=a b c\left(a^{\alpha}+b^{a}+c^{a}\right), \\
g(\alpha)=a^{a+2}(b+c-a)+b^{a+2}(a-b+c)+c^{a+2}(a+b-c) .
\end{array}
$$

Determine the relationship between $f(\alpha)$ and $g(x)$.
(1994 Taipei Mathematical Olympiad Problem)
$$
\text{Solve } \begin{aligned}
f(\alpha)-g(\alpha)= & a b c\left(a^{\alpha}+b^{\alpha}+c^{a}\right)+\left(a^{a+3}+b^{a+3}+c^{a+3}\right) \\
& -a^{a+2}(b+c)-b^{a+2}(a+c)-c^{a+2}(a+b) \\
= & {\left[a^{a+1} b c+a^{a+3}-a^{a+2}(b+c)\right] } \\
& +\left[b^{a+1} a c+b^{a+3}-b^{a+2}(a+c)\right] \\
& +\left[c^{a+1} a b+c^{a+3}-c^{a+2}(a+b)\right] . \\
= & a^{a+1}(a-b)(a-c)+b^{a+1}(b-a)(b-c) \\
& +c^{a+1}(c-a)(c-b) .
\end{aligned}
$$

The right side is exactly the Schur's inequality, hence we know that $f(\alpha)>g(\alpha)$, i.e., $f(\alpha)>g(\alpha)$.