2. Let $a, d$ be non-negative real numbers, $b, c$ be positive real numbers, and $b+c \geqslant a+d$. Then the minimum value of $\frac{b}{c+d}+\frac{c}{a+b}$ is ( ).
(A) 1
(B) $\frac{1}{2}$
(C) $\sqrt{2}-\frac{1}{2}$
(D) $\sqrt{2}$
2. C

Given $b+c \geqslant a+d$, then $b+c \geqslant \frac{1}{2}(a+b+c+d)$. Without loss of generality, assume $a+b \geqslant c+d$, then $\frac{1}{c+d} \geqslant \frac{1}{a+b}$. Therefore,
$$
\begin{array}{l}
\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{c+d}-c\left(\frac{1}{c+d}-\frac{1}{a+b}\right) \\
\geqslant \frac{1}{2} \cdot \frac{a+b+c+d}{c+d}-(c+d)\left[\frac{1}{c+d}-\frac{1}{a+b}\right] \\
=\frac{1}{2} \cdot \frac{a+b}{c+d}+\frac{c+d}{a+b}-\frac{1}{2} \\
\geqslant 2 \cdot \sqrt{\frac{1}{2} \cdot \frac{a+b}{c+d} \cdot \frac{c+d}{a+b}}-\frac{1}{2}=\sqrt{2}-\frac{1}{2} .
\end{array}
$$