设a+b>=c+d.
由于b/(c+d)+c/(a+b)=[(b+c)/(c+d)]-c[1/(c+d)-1/(a+b)]
又因为c=<c+d,及b+c>=a+d,
所以b+c>=1/2*(a+b+c+d)
因此:b/(c+d)+c/(a+b)
>=1/2*[(a+b+c+d)/(c+d)]-(c+d)[1/(c+d)-1/(a+b)]
=1/2*[(a+b)/(c+d)]+[(c+d)/(a+b)]-1/2
>=2×sqrt[(a+b)/(2(c+d))×(c+d)/(a+b)]-1/2
=sqrt(2)-1/2
等号仅当a=sqrt(2)+1,b=sqrt(2)-1,c=2,d=0.
所以b/(c+d)+c/(a+b)最小值为:sqrt(2)-1/2.