10. Non-negative numbers $a$ and $d$, and positive numbers $b$ and $c$ satisfy the condition $b+c \geqslant a+d$. What is the minimum value that the expression $\frac{b}{c+d} + \frac{c}{a+b}$ can take? (1988 22nd All-Soviet Union Mathematical Olympiad)
10. Let's assume $a+b \geqslant c+d$.

Since $\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+b}\right), b+c=\frac{1}{2}(b+c+b+c) \geqslant$ $\frac{1}{2}(a+b+c+d), c \leqslant c+d$, thus $\frac{b}{c+d}+\frac{c}{a+b} \geqslant \frac{1}{2} \cdot \frac{a+b+c+d}{c+d}-(c+$ d) $\left(\frac{1}{c+d}-\frac{1}{a+b}\right)=\frac{1}{2} \cdot \frac{a+b}{c+d}+\frac{c+d}{a+b}-\frac{1}{2} \geqslant 2 \sqrt{\frac{1}{2} \cdot \frac{a+b}{c+d} \cdot \frac{c+d}{a+b}}-\frac{1}{2}=\sqrt{2}$ $-\frac{1}{2}$. When $a=\sqrt{2}+1, b=\sqrt{2}-1, c=2, d=0$, $\frac{b}{c+d} \frac{c}{a+b}$ achieves the minimum value of $\sqrt{2}-$ $\frac{1}{2}$.