6. If $a>b>c>d$, and $\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-d} \geqslant \frac{n}{a-d}$, then the maximum value of $n$ is ( ).
A. 12
B. 9
C. 6
D. 3
$6 . B$
Obviously, $n>0$. From the condition, we get
$$
\begin{aligned}
n & \leqslant \frac{a-d}{a-b}+\frac{a-d}{b-c}+\frac{a-d}{c-d} \\
& =\frac{a-b+b-c+c-d}{a-b}+\frac{a-b+b-c+c-d}{b-c}+\frac{a-b+b-c+c-d}{c-d} \\
& =3+\frac{b-c}{a-b}+\frac{c-d}{a-b}+\frac{a-b}{b-c}+\frac{c-d}{b-c}+\frac{a-b}{c-d}+\frac{b-c}{c-d} \geqslant 9
\end{aligned}
$$