19. Find real numbers $A, B, C$ such that for any real numbers $x, y, z$,
$$A(x-y)(x-z)+B(y-z)(y-x)+C(z-x)(z-y) \geqslant 0$$

is satisfied.

(1988 China National Training Team Selection Examination Problem).
19. In equation (1), let $x=y \neq z$, we get $C(z-x)^{2} \geqslant 0$, thus $C \geqslant 0$. By symmetry, we have
$$A \geqslant 0, B \geqslant 0, C \geqslant 0$$

Let $s=x-y, t=y-z$, then $x-z=s+t$, so (1) is equivalent to
$$A s(s+t)-B s t+C t(s+t) \geqslant 0$$

i.e., $\square$
$$A s^{2}+(A-B+C) s t+C t^{2} \geqslant 0$$

From equation (2), the necessary and sufficient condition for equation (3) to hold for any real numbers $s, t$ is
$$(A-B+C)^{2}-4 A C \leqslant 0$$

i.e., $\square$
$$A^{2}+B^{2}+C^{2} \leqslant 2(A B+B C+C A)$$

Thus, equations (2) and (4) are the required necessary and sufficient conditions.