13 Let $a \geqslant b \geqslant c \geqslant d>0$, try to find the minimum value of the function $f(a, b, c, d)=$ $\left(1+\frac{c}{a+b}\right)\left(1+\frac{d}{b+c}\right)\left(1+\frac{a}{c+d}\right)\left(1+\frac{b}{d+a}\right)$.
Obviously, $f$ has no upper bound, because 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 proves that for any positive numbers $a, b, c, d$ satisfying the conditions, we have $f(a, b, c, d) \geqslant\left(\frac{3}{2}\right)^{4}$, which means we need to prove
$$
\begin{aligned}
& \frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \frac{c+d+a}{3} \cdot \frac{d+a+b}{3} \\
\geqslant & \frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+d}{2} \cdot \frac{d+a}{2} .
\end{aligned}
$$

According to the conditions, $\frac{a}{d} \geqslant \frac{b}{d} \geqslant \frac{c}{d} \geqslant 1$, let
$$
\frac{a}{d}=1+6 x, \frac{b}{d}=1+6 y, \frac{c}{d}=1+6 z,
$$

then $x \geqslant y \geqslant z \geqslant 0$, (1) becomes:
$$
\begin{aligned}
& {[1+2(x+y+z)][1+2(y+z)][1+2(z+x)][1+2(x+y)] } \\
\geqslant & {[1+3(x+y)][1+3(y+z)][1+3 z][1+3 x] . }
\end{aligned}
$$

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

then $\varphi(t)$ and $\psi(t)$ are both fourth-degree 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 (2) holds, we need to prove $p(1) \geqslant 0$, so we just need to prove $A \geqslant 0, B \geqslant 0, C \geqslant 0$. It is easy to see that,
$$
\begin{array}{l} 
A= {\left[8(x+y+z)^{2}+4(x+y)(y+z)\right.} \\
+4(y+z)(z+x)+4(z+x)(x+y)] \\
-[9(x+y)(y+z)+9(y+z)(z+x) \\
+9(z+x)(x+y)+9 x z] \\
= 3\left(x^{2}+y^{2}+z^{2}\right)+x y+y z-8 x z \\
\geqslant 3\left(x^{2}+2 z^{2}\right)-7 x z \geqslant 6 \sqrt{2} x z-7 x z \geqslant 0 . \\
B=\{8(x+y+z)[(x+y)(y+z)+(y+z)(z+x) \\
+(z+x)(x+y)]+8(x+y)(y+z)(z+x)\} \\
- {[27(x+y)(y+z)(z+x)+27 x z(x+z+2 y)] } \\
= 8 x(x+y)(x+z)+8 y(y+x)(y+z) \\
+ 8 z(z+x)(z+y)+5(x+y)(y+z)(z+x) \\
- 27 x z(x+z+2 y)=8\left(x^{3}+y^{3}+z^{3}\right) \\
+ 13\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}\right)-14\left(x^{2} z+x z^{2}\right) \\
- 20 x y z \geqslant 0 .
\end{array}
$$

The above uses
$$
\begin{array}{c}
7\left(x^{3}+y^{3}+z^{3}\right) \geqslant 21 x y z \geqslant 20 x y z+x z^{2}, \\
x^{3} \geqslant \dot{x}^{2} z,
\end{array}
$$

and
$$
\begin{array}{c}
13\left(x^{2} y+x y^{2}\right) \geqslant 13\left(x^{2} z+x z^{2}\right) . \\
c=(x+y)(y+z)[16(x+z)(x+y+z)-81 x z]
\end{array}
$$
$\geqslant 0$.

The above uses
$$
\begin{aligned}
& 16(x+z)(x+y+z) \geqslant 16(x+z)(x+2 z) \\
\geqslant & 16 \cdot 2 \sqrt{x z} \cdot 2 \sqrt{2 x z} \\
= & 64 \sqrt{2} x z \geqslant 81 x z .
\end{aligned}
$$

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