Let $x=a-b, y=b-c, z=c-a, s=a+b+c$, then the inequality can be written as follows:
$$9|s x y z| \leq M\left(s^{2}+x^{2}+y^{2}+z^{2}\right)^{2}$$

where $s, x, y, z$ are any real numbers, and satisfy $x+y+z=0$.
In fact, $s$ is an independent variable. First, we examine the relationship between $x y z$ and $x^{2}+y^{2}+z^{2}$. Since $x+y+z=0$, it is clear that two of $x, y, z$ have the same sign. Without loss of generality, assume $x, y \geq 0$ (the case $x, y \leq 0$ can be similarly proven). By the AM-GM inequality, we have
$$|s x y z|=|\operatorname{sxy}(x+y)| \leq|s| \cdot \frac{(x+y)^{3}}{4}$$

Equality holds when $x=y$. Let $t=x+y$, and apply the AM-GM inequality again, we have
$$2 s^{2} t^{6}=2 s^{2} \cdot t^{2} \cdot t^{2} \cdot t^{2} \leq \frac{\left(2 s^{2}+3 t^{2}\right)^{4}}{4^{4}}$$

Thus,
$$4 \sqrt{2}|s| t^{3} \leq\left(s^{2}+\frac{3}{2} t^{2}\right)^{2} \leq\left(s^{2}+x^{2}+y^{2}+z^{2}\right)^{2}$$

Combining (1) and (2), we get
$$|s x y z| \leq \frac{1}{16 \sqrt{2}}\left(s^{2}+x^{2}+y^{2}+z^{2}\right)^{2}$$

This implies $M \geq \frac{9 \sqrt{2}}{32}$. To prove that $M=\frac{9 \sqrt{2}}{32}$ is the best constant, we need to find $(s, x, y, z)$, i.e., $(a, b, c)$, for which equality holds. After simple calculations, the values for which equality holds are $(a, b, c)=\left(1-\frac{3}{\sqrt{2}}, 1,1+\frac{3}{\sqrt{2}}\right)$.