Example 1.1.16. Determine the least $M$ 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 $a, b$ and $c$.
SOLUTION. Denote \( x = a - b, y = b - c, z = c - a \) and \( s = a + b + c \). Rewrite the inequality in the following form
\[ 9|s x y z| \leq M\left(s^{2} + x^{2} + y^{2} + z^{2}\right)^{2} \]
in which \( s, x, y, z \) are arbitrary real numbers with \( x + y + z = 0 \).
The fact that \( s \) is an independent variable suggests constructing a relationship between \( x y z \) and \( x^{2} + y^{2} + z^{2} \) at first. There are two numbers, say \( x \) and \( y \), with the same sign. Assume that \( x, y \geq 0 \) (the case \( x, y \leq 0 \) is proved similarly). By AM-GM, we have
\[ |s x y z| = |s x y (x + y)| \leq |s| \cdot \frac{(x + y)^{3}}{4} \]
with equality for \( x = y \). Let \( t = x + y \). Applying AM-GM again, we get
\[ 2 s^{2} t^{6} = 2 s^{2} \cdot t^{2} \cdot t^{2} \cdot t^{2} \leq \frac{\left(2 s^{2} + 3 t^{2}\right)^{4}}{4^{4}} \]
and therefore
\[ 4 \sqrt{2} |s| t^{3} \leq \left(s^{2} + \frac{3}{2} t^{2}\right)^{2} \leq \left(s^{2} + x^{2} + y^{2} + z^{2}\right)^{2} \]

Combining (1) and (2), we conclude
\[ |s x y z| \leq \frac{1}{16 \sqrt{2}} \left(s^{2} + x^{2} + y^{2} + z^{2}\right)^{2} \]

This implies that \( M \geq \frac{9 \sqrt{2}}{32} \). To show that \( M = \frac{9 \sqrt{2}}{32} \) is the best constant, we need to find \( (s, x, y, z) \), or in other words, \( (a, b, c) \), for which equality holds. A simple calculation gives \( (a, b, c) = \left(1 - \frac{3}{\sqrt{2}}, 1, 1 + \frac{3}{\sqrt{2}}\right) \).