Example 2 Let non-negative real numbers $a$ and $d$, and positive numbers $b$ and $c$, satisfy the condition $b+c \geqslant a+d$. Find the minimum value of $\frac{b}{c+d}+$ $\frac{c}{a+b}$.
Let's assume $a+b \geqslant c+d$. Since $\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-$ $c\left(\frac{1}{c+d}-\frac{1}{a+b}\right)$, note that $c \leqslant c+d$ and $b+c \geqslant a+d \Leftrightarrow b+c \geqslant \frac{1}{2}(a+b+$ $c+d)$. Therefore, we get
$$\begin{aligned}
\frac{b}{c+d}+\frac{c}{a+b} & \geqslant \frac{1}{2} \frac{a+b+c+d}{c+d}-(c+d)\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \\
& =\frac{1}{2} \frac{a+b}{c+d}+\frac{c+d}{a+b}-\frac{1}{2} \geqslant 2 \sqrt{\frac{a+b}{2(c+d)} \frac{c+d}{a+b}}-\frac{1}{2} \\
& =\sqrt{2}-\frac{1}{2}
\end{aligned}$$

Equality holds if and only if $a=\sqrt{2}+1, b=\sqrt{2}-1, c=2, d=0$, so the minimum value of $\frac{b}{c+d}+$ $\frac{c}{a+b}$ is $\sqrt{2}-\frac{1}{2}$.