求函数y=(x²+8）/(x-1）（x＞1)的最小值

题目

求函数y=(x²+8）/(x-1）（x＞1)的最小值

答案

y=(x^2+8)/(x-1)
=[(x-1)^2+2(x-1)+9]/(x-1)
=(x-1)+9/(x-1)+2
≥2√[(x-1)*9/(x-1)]+2
=2*3+2
=8
当且仅当x-1=9/(x-1),即x=4时取得最小值