5. If $a>b>c>d$, and $\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-d} \geq \frac{n}{a-d}$, then the maximum value of the integer $n$ is $\qquad$

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This problem's difficulty: slightly difficult for the first attempt
Mainly tests: basic inequalities
Solution: 9
$$
\begin{array}{c}
n \leq \frac{a-d}{a-b}+\frac{a-d}{b-c}+\frac{a-d}{c-d} \\
=\frac{a-b+b-c+c-d}{a-b}+\frac{a-b+b-c+c-d}{b-c}+\frac{a-b+b-c+c-d}{c-d} \\
=3+\frac{b-c}{a-b}+\frac{c-d}{a-b}+\frac{a-b}{b-c}+\frac{c-d}{b-c}+\frac{a-b}{c-d}+\frac{b-c}{c-d} \leq 9
\end{array}
$$

By the AM-GM inequality, the above inequality holds, with equality if and only if \(a, b, c, d\) form an arithmetic sequence.