设X,Y属于正实数,xy-(x+y)=1,则x+y最小值xy-(x+1)=1 刚才写错了
问题描述：
设X,Y属于正实数,xy-(x+y)=1,则x+y最小值
xy-(x+1)=1
刚才写错了
答
x+y=xy-1
x+y>=2√xy
xy-1>=2√xy
xy-2√xy-1>=0
x>0,y>0,所以xy>0
所以xy>=1+√2,
xy=(x+y)+1
所以x+y+1>=1+√2
x+y>=√2
所以最小值√2
答
下面写的不错
答
x>0,y>0则x+y>=2(xy)^(1/2)xy-(x+y)=1xy-2(xy)^(1/2)-1>=0解得(xy)^(1/2)=1+2^(1/2)又xy>0xy>=(1+2^(1/2))^2=3+2*2^(1/2)xy-(x+y)=1(x+y)^2-4(x+y)-4>=0x+y>=2+2*2^(1/2)