Question

求函数y=x^2+8/x-1(x>1）的最小值

Answer 1

答：

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

所以：y的最小值为8,此时x-1=9/(x-1),x=4