Determine the smallest number $M$ such that the inequality  
$$ \left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2} $$  
holds for all real numbers $a, b, c$.  (Ireland)
We first consider the cubic polynomial
$$ P(t)=t b\left(t^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c t\left(c^{2}-t^{2}\right). $$
It is easy to check that $P(b)=P(c)=P(-b-c)=0$, and therefore
$$ P(t)=(b-c)(t-b)(t-c)(t+b+c), $$
since the cubic coefficient is $b-c$. The left-hand side of the proposed inequality can therefore be written in the form
$$ \left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right|=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)|. $$
The problem comes down to finding the smallest number $M$ that satisfies the inequality
$$ |(b-c)(a-b)(a-c)(a+b+c)| \leq M \cdot\left(a^{2}+b^{2}+c^{2}\right)^{2}. $$
Note that this expression is symmetric, and we can therefore assume $a \leq b \leq c$ without loss of generality. With this assumption,
$$ |(a-b)(b-c)|=(b-a)(c-b) \leq\left(\frac{(b-a)+(c-b)}{2}\right)^{2}=\frac{(c-a)^{2}}{4} $$
with equality if and only if $b-a=c-b$, i.e., $2 b=a+c$. Also
$$ \left(\frac{(c-b)+(b-a)}{2}\right)^{2} \leq \frac{(c-b)^{2}+(b-a)^{2}}{2} $$
or equivalently,
$$ 3(c-a)^{2} \leq 2 \cdot\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right], $$
again with equality only for $2 b=a+c$. From (2) and (3) we get
$$ \begin{aligned} & |(b-c)(a-b)(a-c)(a+b+c)| \\ \leq & \frac{1}{4} \cdot\left|(c-a)^{3}(a+b+c)\right| \\ = & \frac{1}{4} \cdot \sqrt{(c-a)^{6}(a+b+c)^{2}} \\ \leq & \frac{1}{4} \cdot \sqrt{\left(\frac{2 \cdot\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\right]}{3}\right)^{3} \cdot(a+b+c)^{2}} \\ = & \frac{\sqrt{2}}{2} \cdot\left(\sqrt[4]{\left(\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\right)^{3} \cdot(a+b+c)^{2}}\right)^{2}. \end{aligned} $$
By the weighted AM-GM inequality this estimate continues as follows:
$$ \begin{aligned} & |(b-c)(a-b)(a-c)(a+b+c)| \\ \leq & \frac{\sqrt{2}}{2} \cdot\left(\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\right)^{2} \\ = & \frac{9 \sqrt{2}}{32} \cdot\left(a^{2}+b^{2}+c^{2}\right)^{2}. \end{aligned} $$
We see that the inequality (1) is satisfied for $M=\frac{9}{32} \sqrt{2}$, with equality if and only if $2 b=a+c$ and
$$ \frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2}. $$
Plugging $b=(a+c) / 2$ into the last equation, we bring it to the equivalent form
$$ 2(c-a)^{2}=9(a+c)^{2}. $$
The conditions for equality can now be restated as
$$ 2 b=a+c \quad \text{and} \quad (c-a)^{2}=18 b^{2}. $$
Setting $b=1$ yields $a=1-\frac{3}{2} \sqrt{2}$ and $c=1+\frac{3}{2} \sqrt{2}$. We see that $M=\frac{9}{32} \sqrt{2}$ is indeed the smallest constant satisfying the inequality, with equality for any triple $(a, b, c)$ proportional to $\left(1-\frac{3}{2} \sqrt{2}, 1,1+\frac{3}{2} \sqrt{2}\right)$, up to permutation.

Comment. With the notation $x=b-a, y=c-b, z=a-c, s=a+b+c$ and $r^{2}=a^{2}+b^{2}+c^{2}$, the inequality (1) becomes just $|s x y z| \leq M r^{4}$ (with suitable constraints on $s$ and $r$). The original asymmetric inequality turns into a standard symmetric one; from this point on the solution can be completed in many ways. One can e.g. use the fact that, for fixed values of $\sum x$ and $\sum x^{2}$, the product $x y z$ is a maximum/minimum only if some of $x, y, z$ are equal, thus reducing one degree of freedom, etc. As observed by the proposer, a specific attraction of the problem is that the maximum is attained at a point $(a, b, c)$ with all coordinates distinct.