已知xy是正实数,且xy-x-y=1,求证x+y》2+2根号下2

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已知xy是正实数,且xy-x-y=1,求证x+y》2+2根号下2已知xy是正实数,且xy-x-y=1,求证x+y》2+2根号下2已知xy是正实数,且xy-x-y=1,求证x+y》2+2根号下2x>0,

已知xy是正实数,且xy-x-y=1,求证x+y》2+2根号下2
已知xy是正实数,且xy-x-y=1,求证x+y》2+2根号下2

已知xy是正实数,且xy-x-y=1,求证x+y》2+2根号下2
x>0,y>0
xy-x-y=1
x(y-1)-(y-1)=2
(x-1)(y-1)=2
(x-1)+(y-1)>=2倍根号(x-1)(y-1)>=2倍根号2
x+y>=2+2倍根号2

证明:xy-x-y=1.===>x(y-1)=y+1.===>x=(y+1)/(y-1)=1+[2/(y-1)]===>x=1+[2/(y-1)].且y-1>0.故x+y-2=(y-1)+[2/(y-1)]≥2√2,等号仅当(y-1)²=2时取得。即当x=y=1+√2时取得,故x+y≥2+2√2.