Example 5 Let real numbers $a \geqslant b \geqslant c \geqslant d>0$. Find the minimum value of the function
$$
\begin{array}{l}
f(a, b, c, d) \\
=\left(1+\frac{c}{a+b}\right)\left(1+\frac{d}{b+c}\right)\left(1+\frac{a}{c+d}\right)\left(1+\frac{b}{d+a}\right)
\end{array}
$$
When $a=b=c=d$, the value of $f$ is $\left(\frac{3}{2}\right)^{4}$.
It suffices to prove: For any positive numbers $a, b, c, d$ that satisfy the condition, we have
$$
f(a, b, c, d) \geqslant\left(\frac{3}{2}\right)^{4},
$$

which means 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 condition, 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 $x \geqslant y \geqslant z \geqslant 0$, and inequality (1) becomes
$$
\begin{array}{l}
{[1+2(x+y+z)][1+2(y+z)] .} \\
{[1+2(z+x)][1+2(x+y)]} \\
\geqslant[1+3(x+y)][1+3(y+z)](1+3 z)(1+3 x) .
\end{array}
$$

Reparameterize the constant 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 $P(t)=A t^{2}+B t+C$.
(Here, we use the method of grouping coefficients of the same order to distribute the computational complexity.)
To prove inequality (2), we need to show that $P(1) \geqslant 0$.
It suffices to show that $A \geqslant 0, B \geqslant 0, C \geqslant 0$.
It is easy to see that,
$$
\begin{array}{l} 
A= {\left[8(x+y+z)^{2}+4(x+y)(y+z)+\right.} \\
4(y+z)(z+x)+4(z+x)(x+y)]- \\
{[9(x+y)(y+z)+9(y+z)(z+x)+} \\
9(z+x)(x+y)+9 x z] \\
= 3\left(x^{2}+y^{2}+z^{2}\right)+x y+y z-8 x z \\
\geqslant 3\left(x^{2}+2 z^{2}\right)-7 x z \\
\geqslant 6 \sqrt{2} x z-7 x z \geqslant 0, \\
B=\{8(x+y+z)[(x+y)(y+z)+ \\
(y+z)(z+x)+(z+x)(x+y)]+ \\
8(x+y)(y+z)(z+x)\}- \\
27[(x+y)(y+z)(z+x)+x z(x+z+2 y)] \\
= 8 x(x+y)(x+z)+8 y(y+x)(y+z)+ \\
8 z(z+x)(z+y)+5(x+y)(y+z)(z+x)- \\
27 x z(x+z+2 y) \\
= 8\left(x^{3}+y^{3}+z^{3}\right)+13\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}\right)- \\
14\left(x^{2} z+x z^{2}\right)-20 x y z \\
\geqslant 0,
\end{array}
$$

where we use
$$
\begin{array}{l}
7\left(x^{3}+y^{3}+z^{3}\right) \geqslant 21 x y z \geqslant 20 x y z+x z^{2}, \\
x^{3} \geqslant x^{2} z, \\
13\left(x^{2} y+x y^{2}\right) \geqslant 13\left(x^{2} z+x z^{2}\right) ; \\
C=(x+y)(y+z)[16(x+z)(x+y+z)-81 x z] \\
\geqslant 0,
\end{array}
$$

where we use
$$
\begin{array}{l}
16(x+z)(x+y+z) \geqslant 16(x+z)(x+2 z) \\
\geqslant 16 \times 2 \sqrt{x z} \times 2 \sqrt{2 x z}=64 \sqrt{2} x z \geqslant 81 x z .
\end{array}
$$

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