Question

x+y=1,求4/(x-2)+1/(y+1)的最小值

x+y=1,求4/(x-2)+1/(y+1)的最小值

Answer 1

若x、y∈R+，
则依柯西不等式得
4/(x-2)+1/(y+1)
=2²/(x-2)+1²/(y+1)
≥(2+1)²/[(x-2)+(y+1)]
=9/0
=∞，
故不存在待求的最小值!
请楼主检查题目是否抄错。