1. Let $a, d$ be non-negative real numbers, $b, c$ be positive real numbers, and $b+c \geq a+d$. Then the minimum value of $\frac{b}{c+d}+\frac{c}{a+b}$ is
1. $\sqrt{2}-\frac{1}{2}$

Analysis: Suppose $b+a \geq c+d$. Since $\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+b}\right)$, and $b+c \geq \frac{a+b+c+d}{2}$, therefore
$$
\frac{b}{c+d}+\frac{c}{a+b} \geq \frac{1}{2} \frac{a+b+c+d}{c+d}-(c+d)\left(\frac{1}{c+d}-\frac{1}{a+b}\right)=\frac{1}{2} \frac{a+b}{c+d}+\frac{c+d}{a+b}-\frac{1}{2} \geq \sqrt{2}-\frac{1}{2}
$$

Equality holds if and only if $a=\sqrt{2}+1, b=\sqrt{2}-1, c=2, d=0$.