Let's determine the smallest real number $M$ for which 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 1 Notice that on the left side of the inequality, the expression inside the absolute value changes sign when any two variables are swapped, and thus it is divisible by the difference of any two variables. By factoring, we get that the left side is $|(a-b)(b-c)(c-a)(a+b+c)|$. By introducing the new variables $x=a-b, y=b-c, z=c-a, t=a+b+c$, on one hand, we have $x^{2}+y^{2}+z^{2}+t^{2}=3\left(a^{2}+b^{2}+c^{2}\right)$, and on the other hand, for any set of numbers $x, y, z, t$ such that $x+y+z=0$, there uniquely exist corresponding quantities $a, b, c$. We seek the smallest number $M$ such that

$$
|x y z t| \leq M \frac{\left(x^{2}+y^{2}+z^{2}+t^{2}\right)^{2}}{9}
$$

Among the three new variables, $x, y$, and $z$, there are two with the same sign, which we can assume to be $x$ and $y$. Replacing both with their arithmetic mean increases (or does not decrease) the value on the left side, while the value on the right side decreases (or does not increase). (This follows directly from the inequalities between the geometric, arithmetic, and quadratic means.) Therefore, we can assume that $x=y\left(=-\frac{z}{2}\right)$. Let $u=|x|=|y|=\left|\frac{z}{2}\right|$ and $v=|t|$. With these variables, (1) can be written as

$$
2 u^{3} v \leq M \frac{\left(6 u^{2}+v^{2}\right)^{2}}{9}
$$

Multiplying inequality (2) by $\sqrt{2}$, we get

$$
\sqrt{8} u^{3} v \leq \frac{\sqrt{2}}{9} M\left(6 u^{2}+v^{2}\right)^{2}
$$

which can also be written as:

$$
\sqrt{2} u \cdot \sqrt{2} u \cdot \sqrt{2} u \cdot v \leq \frac{\sqrt{2}}{9} M \cdot 16\left(\frac{(\sqrt{2} u)^{2}+(\sqrt{2} u)^{2}+(\sqrt{2} u)^{2}+v^{2}}{4}\right)^{2}
$$

Taking the fourth root:

$$
\sqrt[4]{\sqrt{2} u \cdot \sqrt{2} u \cdot \sqrt{2} u \cdot v} \leq \sqrt[4]{\frac{16 \sqrt{2}}{9} M} \sqrt{\frac{(\sqrt{2} u)^{2}+(\sqrt{2} u)^{2}+(\sqrt{2} u)^{2}+v^{2}}{4}}
$$

If $M=\frac{9}{16 \sqrt{2}}$, then (3) is precisely the identically satisfied inequality between the geometric and quadratic means, which means that $M \leq \frac{9}{16 \sqrt{2}}$. However, this estimate is sharp, for example, if $a=3+\sqrt{2}, b=\sqrt{2}, c=\sqrt{2}-3$, then equality holds in the problem's inequality. The sought value of $M$ is thus $\frac{9}{16 \sqrt{2}}$.[^0]


[^0]:    ${ }^{1}$ Based on other solutions.