Example 5 Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$, to make the inequality $\log _{x_{1}}^{x_{1}} 1993+\log _{x_{2}}^{x_{1}} 1993+$ $\log _{x_{2}}^{x_{2}} 1993 \geqslant k \log _{x_{1}}^{x_{1}} 1993$ always hold, then the maximum value of $k$ is
Solve by using the change of base formula to transform the inequality in the problem to
$$
\frac{1}{\lg \frac{x_{0}}{x_{1}}}+\frac{1}{\lg \frac{x_{1}}{x_{2}}}+\frac{1}{\lg \frac{x_{2}}{x_{3}}} \geqslant \frac{k}{\lg \frac{x_{0}}{x_{3}}} .
$$

Since $x_{0}>x_{1}>x_{2}>x_{3}>0$, we have $\lg \frac{x_{0}}{x_{1}}>0, \lg \frac{x_{1}}{x_{2}}>0, \lg \frac{x_{2}}{x_{3}}>0, \lg \frac{x_{0}}{x_{3}}>0$.
(1) This is equivalent to (note that $\left.\lg \frac{x_{0}}{x_{1}}+\lg \frac{x_{1}}{x_{2}}+\lg \frac{x_{2}}{x_{3}}=\lg \left(\frac{x_{0}}{x_{1}} \cdot \frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}}\right)=\lg \frac{x_{0}}{x_{3}}\right)$
$$
\left(\frac{1}{\lg \frac{x_{0}}{x_{1}}}+\frac{1}{\lg \frac{x_{1}}{x_{2}}}+\frac{1}{\lg \frac{x_{2}}{x_{3}}}\right)\left(\lg \frac{x_{0}}{x_{1}}+\lg \frac{x_{1}}{x_{2}}+\lg \frac{x_{2}}{x_{3}}\right) \geqslant k .
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\frac{1}{\lg \frac{x_{0}}{x_{1}}}+\frac{1}{\lg \frac{x_{1}}{x_{2}}}+\frac{1}{\lg \frac{x_{2}}{x_{3}}}\right)\left(\lg \frac{x_{0}}{x_{1}}+\lg \frac{x_{1}}{x_{2}}+\lg \frac{x_{2}}{x_{3}}\right) \\
\geqslant\left(\frac{1}{\sqrt{\lg \frac{x_{0}}{x_{1}}}} \cdot \sqrt{\lg \frac{x_{0}}{x_{1}}}+\frac{1}{\sqrt{\lg \frac{x_{1}}{x_{2}}}} \cdot \sqrt{\lg \frac{x_{1}}{x_{2}}}+\frac{1}{\sqrt{\lg \frac{x_{2}}{x_{3}}}} \cdot \sqrt{\lg \frac{x_{2}}{x_{3}}}\right)^{2}=3^{2}=9,
\end{array}
$$

Equality holds if and only if $\lg \frac{x_{0}}{x_{1}}=\lg \frac{x_{1}}{x_{2}}=\lg \frac{x_{2}}{x_{3}}$, i.e., $\frac{x_{0}}{x_{1}}=\frac{x_{1}}{x_{2}}=\frac{x_{2}}{x_{3}}$.
Therefore, for (2) to always hold, it is necessary and sufficient that $k \leqslant 9$, so the maximum value of $k$ is 9.