6・66 $a, b, c$ are positive real numbers, $\alpha$ is a real number, and suppose
$$\begin{array}{l}
f(\alpha)=a b c\left(a^{\alpha}+b^{\alpha}+c^{\alpha}\right) \\
g(\alpha)=a^{\alpha+2}(b+c-a)+b^{\alpha+2}(a-b+c)+c^{\alpha+2}(a+b-c)
\end{array}$$

Determine the relationship in size between $f(\alpha)$ and $g(\alpha)$.
【Solution】 From the given,
$$\begin{aligned}
f(\alpha)-g(\alpha)= & a b c\left(a^{\alpha}+b^{\alpha}+c^{\alpha}\right)+\left(a^{\alpha+3}+b^{\alpha+3}+c^{\alpha+3}\right) \\
& -a^{\alpha+2}(b+c)-b^{\alpha+2}(c+a)-c^{\alpha+2}(a+b) \\
= & {\left[a^{\alpha+1} b c+a^{\alpha+3}-a^{\alpha+2}(b+c)\right] } \\
& +\left[b^{\alpha+1} c a+b^{\alpha+3}-b^{\alpha+2}(c+a)\right] \\
& +\left[c^{\alpha+1} a b+c^{\alpha+3}-c^{\alpha+2}(a+b)\right] \\
= & a^{\alpha+1}(a-b)(a-c)+b^{\alpha+1}(b-c)(b-a)+ \\
& c^{\alpha+1}(c-a)(c-b) .
\end{aligned}$$

We now prove that the right-hand side of the above equation is not less than zero (this is the famous Schur's inequality).
In fact, since $f(\alpha), g(\alpha)$ are symmetric polynomials in $a, b, c$, we may assume without loss of generality that $a \geqslant b \geqslant c$.
Let
$$x=a-b \geqslant 0, y=b-c \geqslant 0 .$$

Substituting into the previous equation, we get
$$\begin{aligned}
f(\alpha)-g(\alpha)= & (b+x)^{\alpha+1} x(x+y)-b^{\alpha+1} x y+(b-y)^{\alpha+1}(-x \\
& -y)(-y) \\
= & (b+x)^{\alpha+1} x(x+y)-b^{\alpha+1} x y+(b-y)^{\alpha+1} y(x \\
& +y)
\end{aligned}$$

Given that $c>0$, we have
$$y-1$ when, $\alpha+1>0$. Noting that $b+x \geqslant b>0$, we have
$$x(x+y)(b+x)^{\alpha+1} \geqslant x y(b+x)^{\alpha+1} \geqslant x y b^{\alpha+1}$$

From (1) and the above inequality, we get
$$f(\alpha) \geqslant g(\alpha), \alpha>-1 \text { when. }$$

When $\alpha=-1$,
$$\begin{aligned}
f(\alpha)-g(\alpha) & =f(-1)-g(-1) \\
& =x(x+y)-x y+y(x+y) \\
& =x^{2}+x y+y^{2} \\
& \geqslant 0
\end{aligned}$$

Therefore,
$$f(\alpha) \geqslant g(\alpha), \alpha=-1 \text { when. }$$

When $\alpha<-1$, $\alpha+1<0$, thus we have
$$(b-y)^{\alpha+1} \geqslant b^{\alpha+1}$$

From this, we get
$$y(x+y)(b-y)^{\alpha+1} \geqslant x y(b-y)^{\alpha+1} \geqslant x y b^{\alpha+1}$$

From (1) and the above inequality, we have
$$f(\alpha) \geqslant g(\alpha), \alpha<-1 \text { when. }$$

In summary, the conclusion of this problem is
$$f(\alpha) \geqslant g(\alpha), \alpha \in R$$