8. Non-negative numbers $a, d$, positive numbers $b, c$, satisfying $b+c \geq a+d$, then the minimum value that $\frac{b}{c+d}+\frac{c}{a+b}$ can take is
Answer: $\sqrt{2}-\frac{1}{2}$
Solution: Without loss of generality, let $a+b \geq c+d$, from the following equation
$$
\begin{array}{c}
\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \\
b+c \geq \frac{1}{2}(a+b+c+d) \\
c \leq c+d
\end{array}
$$

we get
$$
\begin{array}{c}
\frac{b}{c+d}+\frac{c}{a+b} \geq \frac{1}{2} * \frac{a+b+c+d}{c+d}-(c+d)\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \\
=\frac{1}{2} * \frac{(a+b)}{c+d}+\frac{c+d}{a+b}-\frac{1}{2} \\
\geq 2 * \sqrt{\frac{a+b}{2(c+d)} * \frac{c+d}{a+b}}-\frac{1}{2} \\
=\sqrt{2}-\frac{1}{2}
\end{array}
$$

When $a=\sqrt{2}+1, b=\sqrt{2}-1, c=2, d=0$, the above equation holds equality.