题目

设x.y∈R＋,且xy-(x+y)=1,则
x+y≥?或x+y≤?或xy≤?
A.x+y>=2(根号2+1）
x+y

提问时间：2020-10-30

答案

xy-(x+y)=1
xy-x-y=1
得y=（x+1）/(x-1)=1+2/(x-1)
x+y=x+(x+1）/(x-1)=(x-1)+2/(x-1)+2
因为x,y＞0最小值取到时（x-1）=2/(x-1)
既x=（√2）-1
x+y=2√2+2
x+y>=2(xy)^(1/2)
xy-(x+y)=1
xy-2(xy)^(1/2)-1>=0
解得(xy)^(1/2)=1+2^(1/2)
又xy>0
xy>=(1+2^(1/2))^2=3+2*2^(1/2)