9. (16 points) If the quadratic function
$$
f(x)=a x^{2}+b x+c(a, b, c>0)
$$

has zeros, find the maximum value of $\min \left\{\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\right\}$.
Let's take $b=2$.
Given that the quadratic function $f(x)=a x^{2}+b x+c$ has roots, we know $a c \leqslant 1$.
Then $\min \left\{\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}\right\}$
$$
=\min \left\{\frac{2+c}{a}, \frac{c+a}{2}, \frac{a+2}{c}\right\} .
$$

Noting that $a$ and $c$ are symmetric, we can assume $a \leqslant c$.
Thus, $0<a \leqslant 1$, and $\frac{2+c}{a} \geqslant \frac{a+2}{c}$.
Therefore, $\min \left\{\frac{2+c}{a}, \frac{c+a}{2}, \frac{a+2}{c}\right\}$
$$
=\min \left\{\frac{c+a}{2}, \frac{a+2}{c}\right\} \text {. }
$$

Furthermore, when $c \geqslant 2$,
$$
\min \left\{\frac{c+a}{2}, \frac{a+2}{c}\right\}=\frac{a+2}{c} \leqslant \frac{\frac{1}{2}+2}{2}=\frac{5}{4} \text {. }
$$

When $a \leqslant c<2$,
$$
\min \left\{\frac{c+a}{2}, \frac{a+2}{c}\right\}=\frac{c+a}{2},
$$

In this case, if $0<a \leqslant \frac{1}{2}$, then
$$
\frac{c+a}{2}<\frac{2+\frac{1}{2}}{2}=\frac{5}{4} \text {; }
$$

If $\frac{1}{2}<a \leqslant 1$, then
$$
\frac{c+a}{2} \leqslant \frac{a+\frac{1}{a}}{2}<\frac{5}{4} \text {. }
$$

In summary, when $a: b: c=1: 4: 4$ or $4: 4: 1$, the maximum value sought is $\frac{5}{4}$.