9. Given that the real number $m$ satisfies that when the real-coefficient quadratic equation $a x^{2}+b x+c=0$ has real roots,
$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant m a^{2}
$$

always holds. Then the maximum value of $m$ is
9. $\frac{9}{8}$.

Let $\mu=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{a^{2}}$, where $a, b, c$ are real numbers, and $a \neq 0$.

When the equation $a x^{2}+b x+c=0$ has real roots, let the two roots be $x_{1}, x_{2}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a} \text {. } \\
\text { Then } \mu=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{a^{2}} \\
=\left(1-\frac{b}{a}\right)^{2}+\left(\frac{b}{a}-\frac{c}{a}\right)^{2}+\left(\frac{c}{a}-1\right)^{2} \\
=\left(1+x_{1}+x_{2}\right)^{2}+\left(-x_{1}-x_{2}-x_{1} x_{2}\right)^{2}+ \\
\left(x_{1} x_{2}-1\right)^{2} \\
=2\left(x_{1}^{2}+x_{1}+1\right)\left(x_{2}^{2}+x_{2}+1\right) \\
\geqslant 2 \times \frac{3}{4} \times \frac{3}{4}=\frac{9}{8} \text {, } \\
\end{array}
$$

When and only when $x_{1}=x_{2}=-\frac{1}{2}$, i.e., $a=b=4 c \neq 0$, the equality holds.
Therefore, the minimum value of $\mu$ is $\frac{9}{8}$.

Thus, the maximum value of $m$ is $\frac{9}{8}$.