Ta có: $b+c\geq a+d$ $\Leftrightarrow 2(b+c)\geq a+b+c+d$
$\Leftrightarrow b+c\geq \frac{1}{2}(a+b+c+d)$
Mặt khác:
$\frac{b}{c+d}+\frac{c}{a+b}=\frac{b+c}{a+d}-c(\frac{1}{c+d}-\frac{1}{a+b})\geq \frac{1}{2}(\frac{a+b+c+d}{c+d})-(c+d)(\frac{1}{c+d}-\frac{1}{a+b})$
$=\frac{1}{2}(\frac{a+b}{c+d})+\frac{c+d}{a+b}-\frac{1}{2}\geq \sqrt{2}-\frac{1}{2}$
$\rightarrow .......................$