How to find the least real number $M$? Determine the least real number $M$ such that the inequality 
$|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|\le$ $M(a^2+b^2+c^2)^2$ 
holds for all real numbers $a$, $b$ and $c$.

By AM-GM
$$\frac{|\sum\limits_{cyc}ab(a^2-b^2)|}{(a^2+b^2+c^2)^2}=\frac{9|(a-b)(b-c)(c-a)(a+b+c)|}{((a-b)^2+(b-c)^2+(c-a)^2+(a+b+c)^2)^2}\leq$$
$$\leq\frac{9|\frac{(a-b+b-c)^2}{4}(c-a)(a+b+c)|}{\left(\frac{(a-b+b-c)^2}{2}+(c-a)^2+(a+b+c)^2\right)^2}=\frac{9|(c-a)^3(a+b+c)|}{\left(3(c-a)^2+2(a+b+c)^2\right)^2}\leq$$
$$\leq\frac{9|(c-a)^3(a+b+c)|}{\left(4\sqrt[4]{(c-a)^6\cdot2(a+b+c)^2}\right)^2}=\frac{9}{16\sqrt2}.$$
It's obvious that the equality occurs, which says that the answer is $\frac{9}{16\sqrt2}.$
Done!
