11. (20 points) Given positive real numbers $a, b, c, d$ satisfying $a b c d > 1$. Find the minimum value of $\frac{a^{2}+a b^{2}+a b c^{2}+a b c d^{2}}{a b c d-1}$.
11. When $a=b=c=d=2$,
$$
\frac{a^{2}+a b^{2}+a b c^{2}+a b c d^{2}}{a b c d-1}=4 .
$$

By the AM-GM inequality,
$$
\begin{array}{l}
4+a^{2}+2 \cdot \frac{a b^{2}}{2}+4 \cdot \frac{a b c^{2}}{4}+8 \cdot \frac{a b c d^{2}}{8} \\
\geqslant 16 \sqrt[16]{\frac{a^{16} b^{16} c^{16} d^{16}}{2^{32}}}=4 a b c d . \\
\text { Therefore, } \frac{a^{2}+a b^{2}+a b c^{2}+a b c d^{2}}{a b c d-1} \geqslant 4 .
\end{array}
$$

In conclusion, the minimum value sought is 4.