9. (16 points) Let the function $f(x)=\frac{3^{x}}{3^{x}+1}(a, b, c \in$
$\mathbf{R})$, and $a+b+c=0$. Try to determine the relationship between $f(a)+f(b)+f(c)$ and $\frac{3}{2}$.
9. Given the function $f(x)=\frac{3^{x}}{3^{x}+1}$, it satisfies
$$
\begin{array}{l}
f(t)+f(-t)=\frac{3^{t}}{1+3^{t}}+\frac{3^{-t}}{1+3^{-t}} \\
=\frac{3^{t}}{1+3^{t}}+\frac{1}{3^{t}+1}=1, \\
f(0)=\frac{1}{2},
\end{array}
$$

Therefore, when one of $a, b, c$ is zero,
$$
f(a)+f(b)+f(c)=\frac{3}{2} \text {. }
$$

Also, given $a+b+c=0$, then $a, b, c$ must be two positive and one negative or two negative and one positive.
(1) When $a<0, b<0, c>0$,
$$
\begin{array}{l} 
y=3^{-a}>1, z=3^{-b}>1 \Rightarrow y z=3^{c}>1 . \\
\text { Hence } \frac{3}{2}-f(a)-f(b)-f(c) \\
= \frac{3}{2}-(f(-a)+f(a))-(f(-b)+f(b))+ \\
f(-a)+f(-b)-f(c) \\
= f(-a)+f(-b)-f(c)-\frac{1}{2} \\
= \frac{y}{1+y}+\frac{z}{1+z}-\frac{y z}{1+y z}-\frac{1}{2} \\
= \frac{2 y z+y+z}{(1+y)(1+z)}-\frac{3 y z+1}{2(1+y z)} \\
= \frac{y^{2} z^{2}-y z(y+z)+y+z-1}{2(1+y)(1+z)(1+y z)} \\
= \frac{(y z-1)(y-1)(z-1)}{2(1+y)(1+z)(1+y z)}>0 .
\end{array}
$$
(2) When $a>0, b>0, c<0$,
$$
\begin{array}{l}
y=3^{a}>1, z=3^{b}>1 \Rightarrow y z=3^{-c}>1 \text {. }
$$

Hence $f(a)+f(b)+f(c)-\frac{3}{2}$
$$
\begin{array}{l}
=(f(-c)+f(c))+f(a)+f(b)-f(-c)-\frac{3}{2} \\
=f(a)+f(b)-f(-c)-\frac{1}{2} \\
=\frac{y}{1+y}+\frac{z}{1+z}-\frac{y z}{1+y z}-\frac{1}{2}>0 .
\end{array}
$$

Therefore, when one of $a, b, c$ is zero,
$$
f(a)+f(b)+f(c)=\frac{3}{2} \text {; }
$$

When $a, b, c$ are two negative and one positive,
$$
f(a)+f(b)+f(c)>\frac{3}{2} .
$$