11. Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$, to make $\log _{\frac{x_{0}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant$ $k \log _{x_{0}} 1993$ always hold, then the maximum value of $k$ is $\qquad$.
Given that $\log _{\frac{x_{0}}{x_{1}}} 1993, \log _{\frac{x_{1}}{x_{2}}} 1993, \log _{\frac{x_{2}}{x_{3}}} 1993, \log _{\frac{x_{0}}{x_{3}}} 1993>0$,
and $\log _{\frac{x_{1}}{x_{1}}} 1993+\log _{\frac{x_{1}}{x_{1}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant k \log _{\frac{x_{0}}{x_{3}}} 1993$
$\Rightarrow \frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \geqslant \frac{k}{\log _{1993} \frac{x_{0}}{x_{3}}}$
$\Rightarrow k \leqslant \log _{1993} \frac{x_{0}}{x_{3}} \cdot\left(\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}}\right)$.
Notice that $\log _{1993} \frac{x_{0}}{x_{3}}=\log _{1993} \frac{x_{0}}{x_{1}}+\log _{1993} \frac{x_{1}}{x_{2}}+\log _{1993} \frac{x_{2}}{x_{3}}$, so $k \leqslant 9$.