19. (12 分)证明不等式: $a, b, c \in \mathbf{R}, a^{4}+b^{4}+c^{4} \geqslant a b c(a+b+c)$.
$\because a^{4}+b^{4} \geqslant 2 a^{2} b^{2}, \quad b^{4}+c^{4} \geqslant 2 b^{2} c^{2}$,

$c^{4}+a^{4} \geqslant 2 c^{2} a^{2}$,

$\therefore 2\left(a^{4}+b^{4}+c^{4}\right) \geqslant 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)$

即 $a^{4}+b^{4}+c^{4} \geqslant a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}$.

又 $a^{2} b^{2}+b^{2} c^{2} \geqslant 2 a b^{2} c, b^{2} c^{2}+c^{2} a^{2} \geqslant 2 a b c^{2}$,

$c^{2} a^{2}+a^{2} b^{2} \geqslant 2 a^{2} b c$.

$\therefore 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \geqslant 2\left(a b^{2} c+a b c^{2}+a^{2} b c\right)$,

即 $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geqslant a b c(a+b+c)$.

$\therefore a^{4}+b^{4}+c^{4} \geqslant a b c(a+b+c)$.