$9 \cdot 14$ If $m>n>1$, then it must be true that
(A) $\log _{m} n>\log _{n} m>\log _{\frac{1}{n}} m$.
(B) $\log _{n} \frac{1}{m}>\log _{n} m>\log _{m} n$.
(C) $\log _{n} m>\log _{m} n>\log _{n} \frac{1}{n} m$.
(D) $\log _{n} m>\log _{n} \frac{1}{n}>>\log _{m} n$.
(China Beijing High School Mathematics Competition, 1992)
[Solution] $\log _{n} m-\log _{m} n=\log _{n} m-\frac{1}{\log _{n} m}=\frac{\log _{n}^{2} m-1}{\log _{n} m}$, $\because m>n>1, \quad \therefore \quad \log _{n} m>1$,
thus $\log _{n} m-\log _{m} n>0$, which means $\log _{n} m>\log _{m} n$.
$$
\begin{array}{l}
\text { Also } \quad \log _{m} n-\log \frac{1}{n} m=\log _{m} n-\frac{1}{\log _{m} \frac{1}{n}}=\frac{\log _{m}^{2} n+1}{\log _{m} n}, \\
\because m>n>1, \quad \therefore 0<\log _{m} n<1, \quad \therefore \quad \log _{m}^{2} n+1>0, \quad \therefore \quad \frac{\log _{m}^{2} n+1}{\log _{m} n}>0, \quad \therefore \quad \log _{m} n-\log _{n} \frac{1}{n}>0, \quad \therefore \quad \log _{m} n>\log _{n} \frac{1}{n}.
\end{array}
$$
Therefore, the correct choice is (C).