[Algebra]
Problem: 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 $a, b, c$.

Solution: 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

$$
|(b-c)(a-b)(a-c)(a+b+c)| \leq M \cdot\left(a^{2}+b^{2}+c^{2}\right)^{2} . \tag{1}
$$

Note that this expression is symmetric, and we can therefore assume $a \leq b \leq c$ without loss of generality. With this assumption,

$$
|(a-b)(b-c)|=(b-a)(c-b) \leq\left(\frac{(b-a)+(c-b)}{2}\right)^{2}=\frac{(c-a)^{2}}{4} \tag{2}
$$

with equality if and only if $b-a=c-b$, i.e. $2 b=a+c$. Also

$$
\left(\frac{(c-b)+(b-a)}{2}\right)^{2} \leq \frac{(c-b)^{2}+(b-a)^{2}}{2} \tag{3}
$$

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.



Final answer: $M=\frac{9}{32} \sqrt{2}$