11. Let any real numbers $a>b>c>d>0$, to make $\log _{\frac{b}{a}} 2004+\log _{\frac{b}{s}} 2004+\log _{\frac{d}{c}} 2004 \geqslant m \log _{\frac{d}{a}} 2004$ always hold, then the minimum value of $m$ is $\qquad$ .
11. Let $x_{1}=-\log _{2004} \frac{b}{a}, x_{2}=-\log _{2004} \frac{c}{b}, x_{3}=-\log _{2004} \frac{d}{c}$, since $a>b>c>d>0$, we have $x_{1}>0$, $x_{2}>0, x_{3}>0$, the original expression can be transformed into $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}} \leqslant m \cdot \frac{1}{x_{1}+x_{2}+x_{3}}$,
which means $m \geqslant\left(x_{1}+x_{2}+x_{3}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right) \geqslant 9$.
When $x_{1}=x_{2}=x_{3}$, $a, b, c, d$ form a geometric sequence, and the equality holds, so the minimum value of $m$ is 9, hence the answer is 9.