Determine the smallest real number $M$ such that the inequality $\mid a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-\right.$ $\left.a^{2}\right) \mid \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2}$ holds for any real numbers $a, b, c$.

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Author: Kanel-Belov A.Y.

The left side turns to zero when $a=b$, which means it is divisible by $a-b$. Similarly, it is divisible by $a-c$ and $b-c$. Now, from symmetry, it is clear that $a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)=\alpha(a-b)(a-c)(b-c)(a+b+c)$. By comparing the coefficients of $a^{3}$, we find $\alpha=1$. Thus, the problem can be restated as:

find the smallest $M$ for which the inequality $|(a-b)(a-c)(b-c)(a+b+c)| \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2}$ holds. $(*)$

Due to symmetry, without loss of generality, we can assume that $a \leq b \leq c$. Let $c-a=2 t,|a+b+c|=35$.

We have $(b-a)(c-b) \leq 1 / 4((b-a)+(c-b))^{2}=t^{2},\left((b-a)^{2}+(c-b)^{2}\right) \geq 1 / 2((b-a)+(c-b))^{2}=2 t^{2}$. Therefore,

$$
|(a-b)(a-c)(b-c)(a+b+c)| \leq 6 t^{3} s=\frac{1}{\sqrt{2}} \cdot\left(\sqrt[4]{2 t^{2} \cdot 2 t^{2} \cdot 2 t^{2} \cdot 9 s^{2}}\right)^{2} \leq
$$

$\frac{1}{\sqrt{2}}\left(\frac{2 t^{2}+2 t^{2}+2 t^{2}+9 s^{2}}{4}\right)^{2}=$

$=\frac{\sqrt{2}}{32}\left(6 t^{2}+9 s^{2}\right)^{2} \leq \frac{\sqrt{2}}{32}\left((a-b)^{2}+(a-c)^{2}+(b-c)^{2}+(a+b+c)^{2}\right)^{2}=$

$\frac{\sqrt{2}}{32}$

$\left(3 a^{2}+3 b^{2}+3 c^{2}\right)^{2}$

Thus, inequality $(*)$ holds for $M=\frac{9 \sqrt{2}}{32}$. Moreover, equality can be achieved if all inequalities in the estimates turn into equalities (i.e., if $b-a=c-b=t$ and $2 t^{2}=9 s^{2}$). This will happen, for example, when $b=1, s=3, t=\frac{3 \sqrt{2}}{2}, a=1-\frac{3 \sqrt{2}}{2}, c=1+\frac{3 \sqrt{2}}{2}$. Therefore, it is impossible to reduce $M$.

## Answer

$M=\frac{9 \sqrt{2}}{32}$.

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