若正实数x,y满足2x+y+6=xy,x+y的最小值是
问题描述：
若正实数x,y满足2x+y+6=xy,x+y的最小值是
答
2x+y+6=xy
y=(2x+6)/(x-1) >0
∴ x>1
x+y
=x+(2x+6)/(x-1)
=x+[(2x-2)+8]/(x-1)
=x+2+8/(x-1)
=(x-1)+8/(x-1)+3
≥2√8 +3
=4√2 +3
当且仅当 x-1=8/(x-1),即 x=2√2+1时等号成立
所以 x+y的最小值是4√2 +3