a=(b+1)/(b-1)>0,b>1
a+b=(b+1)/(b-1)+b=(b^2+1)/(b-1)
令s=b-1
b^2=[c+1]^2=c^2+2c+1
a+b=(c^2+2c+2)/c
=c+2/c+2=f(c)
f'(c)=1-2/c^2=0,c=√2
f(c)=2+2√2
这是a+b的最小值