3 Find the smallest real number $M$, such that for all real numbers $a, b$ and $c$, we have
$$
\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} .
$$
First, consider
$$
P(t)=t b\left(t^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c t\left(c^{2}-t^{2}\right),
$$

It is easy to see that,
$$
P(b)=P(c)=P(-c-b)=0,
$$

Therefore, we have
$$
\begin{aligned}
& \left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \\
= & |P(a)|=|(b-c)(a-b)(a-c)(a+b+c)|,
\end{aligned}
$$

The original inequality is equivalent to
$$
|(b-c)(a-b)(a-c)(a+b+c)| \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2} .
$$

Due to symmetry, assume without loss of generality that $a \leqslant b \leqslant c$. Then we have
$$
\begin{aligned}
|(a-b)(b-c)| & =(b-a)(c-b) \leqslant\left(\frac{(b-a)+(c-b)}{2}\right)^{2} \\
& =\frac{(c-a)^{2}}{4},
\end{aligned}
$$

and the equality holds if and only if $b-a=c-b$, i.e., $2 b=a+c$. Also,
$$
\left(\frac{(c-b)+(b-a)}{2}\right)^{2} \leqslant \frac{(c-b)^{2}+(b-a)^{2}}{2},
$$

or equivalently,
$$
3(c-a)^{2} \leqslant 2\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right],
$$

and the equality holds if and only if $2 b=a+c$. Therefore,
$$
\begin{aligned}
\mid(b & -c) \left.(a-b)(a-c)(a+b+c)\left|\leqslant \frac{1}{4}\right|(c-a)^{3}(a+b+c) \right\rvert\, \\
& =\frac{1}{4} \sqrt{(c-a)^{6}(a+b+c)^{2}} \\
& \leqslant \frac{1}{4} \sqrt{\left(\frac{2\left((b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right)}{3}\right)^{3}(a+b+c)^{2}} \\
& \leqslant \frac{\sqrt{2}}{2}\left(\sqrt[4]{\left(\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\right)^{3}(a+b+c)^{2}}\right)^{2}
\end{aligned}
$$

By the weighted $A M-G M$ inequality, we have
$$
\begin{aligned}
& |(b-c)(a-b)(a-c)(a+b+c)| \\
\leqslant & \frac{\sqrt{2}}{2}\left(\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\right)^{2} \\
= & \frac{9 \sqrt{2}}{32}\left(a^{2}+b^{2}+c^{2}\right)^{2} .
\end{aligned}
$$

Thus, $M=\frac{9 \sqrt{2}}{32}$, and the equality holds if and only if $2 b=a+c$, and
$$
\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2} .
$$

Solving, we get $2 b=a+c,(c-a)^{2}=18 b^{2}$. Taking $b=1$, we obtain $a=1-\frac{3}{2} \sqrt{2}$ and $c=1+\frac{3}{2} \sqrt{2}$. Therefore, when $\left(1-\frac{3}{2} \sqrt{2}, 1, 1+\frac{3}{2} \sqrt{2}\right)$, the equality holds, hence $M=\frac{9}{32} \sqrt{2}$.