5. Let 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
5. Rewrite the original inequality as
$$
\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \geqslant k \frac{1}{\log _{1993} \frac{x_{0}}{x_{1}} \cdot \frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}}},
$$

i.e.,
$$
\begin{array}{c}
\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \\
\geqslant k \frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}+\log _{1993} \frac{x_{1}}{x_{2}}+\log _{1993} \frac{x_{2}}{x_{3}}} . \\
\text { Let } t_{1}=\log _{1993} \frac{x_{0}}{x_{1}}, t_{2}=\log _{1993} \frac{x_{1}}{x_{2}}, t_{3}=\log _{1993} \frac{x_{2}}{x_{3}}
\end{array}
$$

Then the original inequality becomes
$$
\left(t_{1}+t_{2}+t_{3}\right)\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}+\frac{1}{t_{3}}\right) \geqslant k .
$$

However,
$$
\begin{aligned}
& \left(t_{1}+t_{2}+t_{3}\right)\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}+\frac{1}{t_{3}}\right) \\
\geqslant & 3 \sqrt[3]{t_{1} \cdot t_{2} \cdot t_{3}} \cdot 3 \sqrt[3]{\frac{1}{t_{1}} \cdot \frac{1}{t_{2}} \cdot \frac{1}{t_{3}}}=9,
\end{aligned}
$$

with equality holding if and only if \( t_{1}=t_{2}=t_{3} \) (i.e., \( x_{0}, x_{1}, x_{2}, x_{3} \) form a geometric sequence), \(\therefore \quad k_{\max }=9\).