II. (50 points) Let real numbers $a \geqslant b \geqslant c \geqslant d>0$. Find the minimum value of the function
$$
\begin{array}{l}
f(a, b, c, d) \\
=\left(1+\frac{c}{a+b}\right)\left(1+\frac{d}{b+c}\right)\left(1+\frac{a}{c+d}\right)\left(1+\frac{b}{d+a}\right)
\end{array}
$$
Obviously, $f$ has no upper bound, which is due to when $a=b=1$, $c+d \rightarrow 0$, $f \rightarrow+\infty$. Also note that $f$ is a zero-degree homogeneous function, and when $a=b=c=d$, the value of $f$ is $\left(\frac{3}{2}\right)^{4}$.

The following proof shows that for any positive numbers $a, b, c, d$ satisfying the conditions, we have $f(a, b, c, d) \geqslant\left(\frac{3}{2}\right)^{4}$, i.e., we need to prove
$$
\begin{array}{l}
\frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \frac{c+d+a}{3} \cdot \frac{d+a+b}{3} \\
\geqslant \frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+d}{2} \cdot \frac{d+a}{2} .
\end{array}
$$

According to the conditions, we have $\frac{a}{d} \geqslant \frac{b}{d} \geqslant \frac{c}{d} \geqslant 1$.
Let $\frac{a}{d}=1+6 x, \frac{b}{d}=1+6 y, \frac{c}{d}=1+6 z$, then $x \geqslant y \geqslant z \geqslant 0$.
Thus, equation (1) becomes
$$
\begin{array}{l}
\quad[1+2(x+y+z)][1+2(y+z)][1+2(z+x)] . \\
\quad[1+2(x+y)] \\
\geqslant[1+3(x+y)][1+3(y+z)](1+3 z)(1+3 x) . \text { (2) }
\end{array}
$$

Activate a constant, and denote 1 as $t$, and set
$$
\begin{aligned}
\varphi(t)= & {[t+2(x+y+z)][t+2(y+z)] . } \\
& {[t+2(z+x)][t+2(x+y)], } \\
\psi(t)= & {[t+3(x+y)][t+3(y+z)](t+3 z) . } \\
& (t+3 x) .
\end{aligned}
$$

Then $\varphi(t)$ and $\psi(t)$ are both quartic polynomials in $t$. And $p(t)=$ $\varphi(t)-\psi(t)$ is a quadratic polynomial in $t$.

Let $p(t)=A t^{2}+B t+C$. To prove that equation (2) holds, we need to prove that $p(1) \geqslant 0$.
Thus, we only need to prove that $A \geqslant 0, B \geqslant 0, C \geqslant 0$.
It is easy to see that
$$
\begin{array}{l} 
A= {\left[8(x+y+z)^{2}+4(x+y)(y+z)+\right.} \\
4(y+z)(z+x)+4(z+x)(x+y)]- \\
{[9(x+y)(y+z)+9(y+z)(z+x)+} \\
9(z+x)(x+y)+9 x z] \\
=3\left(x^{2}+y^{2}+z^{2}\right)+x y+y z-8 x z \\
\geqslant 3\left(x^{2}+2 z^{2}\right)-7 x z \\
\geqslant 6 \sqrt{2} x z-7 x z \geqslant 0, \\
B=\{8(x+y+z)[(x+y)(y+z)+ \\
(y+z)(z+x)+(z+x)(x+y)]+ \\
8(x+y)(y+z)(z+x)\}- \\
{[27(x+y)(y+z)(z+x)+27 x z(x+z+2 y)] } \\
= 8 x(x+y)(x+z)+8 y(y+x)(y+z)+ \\
8 z(z+x)(z+y)+5(x+y)(y+z)(z+x)- \\
27 x z(x+z+2 y) \\
= 8\left(x^{3}+y^{3}+z^{3}\right)+13\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}\right)- \\
14\left(x^{2} z+x z^{2}\right)-20 x y z \\
\end{array}
$$
$\geqslant 0$,
(because 7 $\left(x^{3}+y^{3}+z^{3}\right) \geqslant 21 x y z \geqslant 20 x y z+x z^{2}, x^{3} \geqslant x^{2} z$,
$$
\begin{array}{l}
\left.13\left(x^{2} y+x y^{2}\right) \geqslant 13\left(x^{2} z+x z^{2}\right)\right) \\
\quad C=(x+y)(y+z)[16(x+z)(x+y+z)-81 x z]
\end{array}
$$
$$
\geqslant 0 \text {. }
$$
(because $16(x+z)(x+y+z) \geqslant 16(x+z)(x+2 z)$
$$
\geqslant 16 \times 2 \sqrt{x z} \times 2 \sqrt{2 x z}=64 \sqrt{2} x z \geqslant 81 x z)
$$

Therefore, $p(1) \geqslant 0$.
Thus, the minimum value of the function $f(a, b, c, d)$ is $\left(\frac{3}{2}\right)^{4}$.