II Let $a, d \geqslant 0, b, c>0$ and $b+c \geqslant a+d$. Find the minimum value of $\frac{b}{c+d}+\frac{c}{a+b}$.
11. From the given, we have $b+c \geqslant \frac{1}{2}(a+b+c+d)$. Without loss of generality, assume $a+b=c+d$, then
$$\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}+c\left(\frac{1}{a+b}-\frac{1}{c+d}\right) \geqslant \frac{\frac{1}{2}(a+b+c+d)}{c+d}+(c+$$
d) $\left(\frac{1}{a+b}-\frac{1}{c+d}\right)=\frac{a+b}{2(c+d)}+\frac{c+d}{a+b}-\frac{1}{2} \geqslant \sqrt{2}-\frac{1}{2}$. When $a=\sqrt{2}+1, b=$ $\sqrt{2}-1, c=2, d=0$, the equality holds. Therefore, the minimum value sought is $\sqrt{2}-\frac{1}{2}$.