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

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 $c>0$, 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} \text {, }
$$

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
$$
(b-y)^{\alpha+1} \geqslant b^{\alpha+1} \text {, }
$$

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
$$
f(\alpha) \geqslant g(\alpha), \alpha \in R
$$