6. (2000 "Hope Cup" Invitational Competition Question) Let $a>b>c$, and $\frac{1}{a-b}+\frac{1}{b-c} \geqslant \frac{n}{a-c}$ always holds, then what is the maximum value of $n$?
$$
\begin{array}{l}
\text { 6. Let } \vec{a}=\left(\frac{1}{\sqrt{a-b}}, \frac{1}{\sqrt{b-c}}\right), \vec{b}=(\sqrt{a-b}, \sqrt{b-c}), \\
\frac{1}{a-b}+\frac{1}{b-c}=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(1+1)^{2}}{(a-b)+(b-c)}=\frac{4}{a-c} .
\end{array}
$$

Then the maximum value of $n$ is 4.