Problem: Determine the least real number $M$ such that the inequality
\[ \left\lvert ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) \right\rvert
  \leq M\left( a^2+b^2+c^2 \right)^2 \]
holds for all real numbers $a$, $b$ and $c$.

Solution: It's the same as
\[ \left\lvert (a-b)(b-c)(c-a)(a+b+c) \right\rvert
\le M \left( a^2+b^2+c^2 \right)^2. \]
Let $x=a-b$, $y=b-c$, $z=c-a$, $s=a+b+c$.
Then we want to have
\[
  \left\lvert xyzs \right\rvert
  \le \frac{M}{9} (x^2+y^2+z^2+s^2)^2.
\]
Here $x+y+z=0$.

Now if $x$ and $y$ have the same sign,
we can replace them with the average
(this increases the LHS and decreases RHS).
So we can have $x=y$, $z=-2x$.
Now WLOG $x > 0$ to get
\[ 2x^3 \cdot s \le \frac{M}{9} \left( 6x^2+s^2 \right)^2. \]
After this routine calculation gives
$M = \frac{9}{32}\sqrt2$ works and is optimal
(by $6x^2+s^2 = 2x^2 + 2x^2 + 2x^2 + s^2$ and AM-GM).