Example 5 Let $a$ and $d$ be non-negative real numbers, $b$ and $c$ be positive numbers, and $b+c \geqslant a+d$. Find the minimum value of $\frac{b}{c+d}+$ $\frac{c}{a+b}$.
Solution: Without loss of generality, let $b+c=1$ (otherwise, multiply a ratio on $a, b, c, d$ simultaneously), then $a+b+c+d \leqslant 2(b+c)=2$. Among $a+b$ and $c+d$, there must be one not greater than 1. Without loss of generality, let $a+b \leqslant 1$. Let $u=a+b$, then $\frac{b}{c+d}+\frac{c}{a+b} \geqslant \frac{b}{2-u}+\frac{c}{u}=\frac{b}{2-u}+\frac{1-b}{u}=\frac{1}{u}+\left(\frac{1}{2-u}-\frac{1}{u}\right) \cdot b$. Since $\frac{1}{2-u} \leqslant 1 \leqslant \frac{1}{u}, b \leqslant a+b=u$, so $\frac{b}{c+d}+\frac{c}{a+b} \geqslant \frac{1}{u}+\left(\frac{1}{2-u}-\frac{1}{u}\right) u=\frac{1}{u}+\frac{u}{2-u}-1=\frac{2-u}{2 u}+\frac{u}{2-u}-\frac{1}{2} \geqslant 2 \sqrt{\frac{2-u}{2 u} \cdot \frac{u}{2-u}}-\frac{1}{2}=\sqrt{2}-\frac{1}{2}$, when $a=0, b=2 \sqrt{2}-2, c=3-2 \sqrt{2}, d=1$ the equality holds. Therefore, the minimum value sought is $\sqrt{2}-\frac{1}{2}$.