设X,Y属于正实数,xy-(x+y)=1,则x+y最小值
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>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)