Question

当x>1时，不等式x+1/(x-1)≥a恒成立，则实数a的范围是?

Answer 1

a应小于或等于x+1/(x-1)的最小值
由均值不等式得，x+1/(x-1)的最小值为1
所以取值范围得a小于等于3

Answer 2

x+1/(x-1)=(x-1)+1/(x-1)+1≥2+1=3
所以a<=3

Answer 3

f[x]=x+1/(x-1),f[x]的导数为[(x-1)^2-1]/(x-1)^2
x=2时f[x]取最小值f[2]=5/2,只要保证最小值大于等于a就成立得a<=5/2.

Answer 4

当x>1时 1+[(x-1)+1/(x-1)]≥1+2 所以a<=3

Answer 5

设X-1=T （T>0)
左式
=1/T+T+1
≥2（√1/T*T）+1
=2+1
=3
故a≤3

Answer 6

a<=3