6.96 Let $a$ and $d$ be non-negative numbers, $b$ and $c$ be positive numbers, and $b+c \geqslant a+d$. Find the minimum value of the following expression:
$$
\frac{b}{c+d}+\frac{c}{a+b} \text {. }
$$
[Solution] If we simultaneously swap $a$ with $d$ and $b$ with $c$, the conditions and conclusions of the problem remain unchanged. Therefore, we can assume without loss of generality that $a+b \geqslant c+d$. Thus, we have
$$
\begin{aligned}
& \frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \\
\geqslant & \frac{\frac{1}{2}(a+b+c+d)}{c+d}-(c+d)\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \\
= & \frac{a+b}{2(c+d)}-\frac{1}{2}+\frac{c+d}{a+b} \\
\geqslant & 2 \cdot \sqrt{\frac{a+b}{2(c+d)} \cdot \frac{c+d}{a+b}}-\frac{1}{2} \\
= & \sqrt{2}-\frac{1}{2} .
\end{aligned}
$$

On the other hand, when $a=\sqrt{2}+1, b=\sqrt{2}-1, c=2, d=0$, we have
$$
\frac{b}{c+d}+\frac{c}{a+b}=\sqrt{2}-\frac{1}{2} \text {. }
$$

Therefore, the minimum value sought is $\sqrt{2}-\frac{1}{2}$.