设x,y均为正实数,且 xy=x+y+8,则xy的最小值为?
x>0,y>0,且xy=x+y+8
xy=x+y+8≥2√xy+8
xy-2√xy+8≥0
(√xy+2)(√xy-4)≥0
√xy≤-2====>xy≤4
√xy≥4=====>xy≥16
xy的取值范围是xy=16