Prove that there obviously exist positive numbers $x, y, z$ such that $a=y+z, b=z+x, c=x+y$. Since
$$\begin{aligned} & a^{2} b(a-b)=(y+z)^{2}(z+x)(y-x) \\ = & (y+z)(z+x)\left(y^{2}-z^{2}\right)+(y+z)^{2}\left(z^{2}-x^{2}\right) \end{aligned}$$

Similarly handling $b^{2} c(b-c), c^{2} a(c-a)$, we have
$$\begin{aligned} & a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \\ = & 2 x(y-z) y^{2}+2 y(z-x) z^{2}+2 z(x-y) x^{2} \end{aligned}$$

The original inequality is equivalent to
$$x y z(x+y+z) \leqslant x y^{3}+y z^{3}+z x^{3}$$

By the Cauchy-Schwarz inequality, we get
$$\begin{array}{c} x+y+z \leqslant\left(\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x}\right)^{\frac{1}{2}}(x+y+z)^{\frac{1}{2}} \\ x+y+z \leqslant \frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x} \end{array}$$

Thus, the original inequality holds, and equality holds when $a=b=c$.