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

Let $a, d$ be non-negative numbers, $b, c$ be positive numbers, and $b+c \geq a+d$. Then the minimum value of $\frac{b}{c+d}+\frac{c}{a+b}$ is
Solve: $\sqrt{2}-\frac{1}{2}$.
If $a, d$ are swapped and $b, c$ are swapped simultaneously, the conditions and the objective function in the problem remain unchanged, so we may assume
$$
a+b \geq c+d
$$

Then
$$
\begin{aligned}
\frac{b}{c+d}+\frac{c}{a+b}= & \frac{b+c}{c+d}+c\left(\frac{1}{a+b}-\frac{1}{c+d}\right) \geq \frac{2(b+c)}{2(c+d)}+(c+d)\left(\frac{1}{a+b}-\frac{1}{c+d}\right) \\
& \geq \frac{b+c+a+d}{2(c+d)}+\frac{c+d}{a+b}-1=\frac{a+b}{2(c+d)}+\frac{c+d}{a+b}-\frac{1}{2} \\
& \geq 2 \sqrt{\frac{a+b}{2(c+d)} \cdot \frac{c+d}{a+b}}-\frac{1}{2} \geq \sqrt{2}-\frac{1}{2} .
\end{aligned}
$$