Question

设X,Y属于正实数,xy-(x+y)=1,则x+y最小值

Answer 1

x>0,y>0

则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)

xy-(x+y)=1

(x+y)^2-4(x+y)-4>=0

x+y>=2+2*2^(1/2)