Example 11 If any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>$ 0 , to make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is $\qquad$
(1993, National High School Mathematics Competition)
Analysis: The inequality can be transformed into
$$
\begin{array}{l}
k \leqslant-\frac{\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993}{\log _{\frac{x_{0}}{x_{3}}} 1993} \\
=f\left(x_{0}, x_{1}, x_{2}, x_{3}\right) \text {. } \\
\end{array}
$$

Thus,
$$
\{k\}_{\text {max }}=\left\{f\left(x_{0}, x_{1}, x_{2}, x_{3}\right)\right\}_{\text {min }} .
$$

By the harmonic mean inequality, we get
$$
\begin{array}{l}
f\left(x_{0}, x_{1}, x_{2}, x_{3}\right) \geqslant \\
\frac{3}{\left(\log _{1993} \frac{x_{0}}{x_{1}}+\log _{1993} \frac{x_{1}}{x_{2}}+\log _{1993} \frac{x_{2}}{x_{3}}\right) \log _{\frac{x_{0}}{x_{3}}} 1993} \\
\end{array}
$$
$$
=3 \text {. }
$$

Therefore, $k_{\max }=3$.