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
人气:458 ℃ 时间:2020-04-13 02:56:01
解答
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)
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