12. $a>b>c, n \in \mathbf{N}$ 且 $\frac{1}{a-b}+\frac{1}{b-c} \geqslant \frac{n}{a-c}$, 求 $n$ 的最大值.
$\because a>b>c, \therefore a-b>0, b-c>0, a-c>0$.

$\because \frac{1}{a-b}+\frac{1}{b-c} \geqslant \frac{n}{a-c}$,

$\therefore n \leqslant \frac{a-c}{a-b}+\frac{a-c}{b-c}$.

$\because a-c=(a-b)+(b-c)$,

$\therefore n \leqslant \frac{(a-b)+(b-c)}{a-b}+\frac{(a-b)+(b-c)}{b-c}$,

$\therefore n \leqslant \frac{b-c}{a-b}+\frac{a-b}{b-c}+2$.

$\because \frac{b-c}{a-b}+\frac{a-b}{b-c} \geqslant 2 \sqrt{\left(\frac{b-c}{a-b}\right) \cdot\left(\frac{a-b}{b-c}\right)}$

$=2(2 b=a+c$ 时取等号 $)$.

$\therefore n \leqslant 4 . \therefore n$ 的最大值是 4 .