5. 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:
$$
\frac{b}{c+d}+\frac{c}{a+b} .
$$
(22nd All-Soviet Union Olympiad)
5. If $a$ and $d$ are swapped, and $b$ and $c$ are swapped simultaneously, neither the given conditions nor the conclusion changes. Therefore, without loss of generality, we can assume $a+b \geqslant c+d$. Thus, we have 
\[
\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \geqslant \frac{\frac{1}{2}(a+b+c+d)}{c+d}-(c+d)\left(\frac{1}{c+d}-\frac{1}{a+b}\right)=\frac{a+b}{2(c+d)}-\frac{1}{2}+\frac{c+d}{a+b} \geqslant 2 \sqrt{\frac{a+b}{2(c+d)} \cdot \frac{c+d}{a+b}}-\frac{1}{2}=\sqrt{2}-\frac{1}{2}.
\]

On the other hand, when $a=\sqrt{2}+1, b=\sqrt{2}-1, c=2, d=0$, we have 
\[
\frac{b}{c+d}+\frac{c}{a+b}=\sqrt{2}-\frac{1}{2}.
\]
Therefore, the minimum value sought is $\sqrt{2}-\frac{1}{2}$.