Question

设正数x,y满足xy-x-2y＝0,x>0,则x+y最小值

Answer 1

设t=x+y

则：x=t-y

(t-y)y-(t-y)-2y

=-y^2+(t-1)y-t=0

判别式△=(t-1)^2-4t=t^2-6t+1≥0

t≤3-2√2,或,t≥3+2√2

y1y2=t>0

y1+y2=t-1>0,t>1

所以,t≥3+2√2

即：x+y最小值为：3+2√2