11. (20 points) Let positive real numbers \(a, b, c, d \in \left[\frac{1}{2}, 2\right]\), and satisfy \(a b c d = 1\). Try to find
\[
\begin{array}{l}
f(a, b, c, d) \\
=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{d}\right)\left(d+\frac{1}{a}\right)
\end{array}
\]

the maximum and minimum values.
11. Given $a b c d=1$, we have
$$
\begin{array}{l}
f(a, b, c, d) \\
=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{d}\right)\left(d+\frac{1}{a}\right) \\
=(a b+1)(b c+1)(c d+1)(d a+1) .
\end{array}
$$

By the AM-GM inequality, we get
$$
\begin{array}{l}
(a b+1)(c d+1)=2+a b+c d \\
\geqslant 2+2 \sqrt{a b c d}=4 .
\end{array}
$$

Similarly, $(b c+1)(d a+1) \geqslant 4$.
$$
\text { Therefore, }(a b+1)(b c+1)(c d+1)(d a+1) \geqslant 16 \text {, }
$$

with equality if and only if
$$
a=c=t, b=d=\frac{1}{t}\left(t \in\left[\frac{1}{2}, 2\right]\right)
$$

Given $\frac{1}{2} \leqslant a, c \leqslant 2$, we know $c \leqslant 4 a, a \leqslant 4 c$.
Thus, $(4 a-c)(4 c-a) \geqslant 0$
$$
\Rightarrow a^{2}+c^{2} \leqslant \frac{17}{4} a c \text {. }
$$

Hence, $\frac{a}{c}+\frac{c}{a} \leqslant \frac{17}{4}$, and $(a+c)^{2} \leqslant \frac{25}{4} a c$, which means
$$
a+c \leqslant \frac{5}{2} \sqrt{a c} \text {. }
$$

Similarly, $\frac{b}{d}+\frac{d}{b} \leqslant \frac{17}{4} \Rightarrow b+d \leqslant \frac{5}{2} \sqrt{b d}$.
Therefore, $(a+c)(b+d) \leqslant \frac{25}{4} \sqrt{a b c d}=\frac{25}{4}$.
Thus, $f(a, b, c, d)$
$$
\begin{array}{l}
=\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{d}\right)\left(d+\frac{1}{a}\right) \\
=4+\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}+2(a+c)(b+d) \\
\leqslant 4+\frac{17}{4}+\frac{17}{4}+2 \times \frac{25}{4}=25 . \\
\text { When }(a, b, c, d)=\left(2,2, \frac{1}{2}, \frac{1}{2}\right) \text {, the equality holds. }
\end{array}
$$

When $(a, b, c, d)=\left(2,2, \frac{1}{2}, \frac{1}{2}\right)$, the equality holds.

In summary, the minimum and maximum values of $f(a, b, c, d)$ are 16 and 25, respectively.