Question

设a>b>c,则n是整数,且1/(a-b)+1/(b-c)>=n/(a-c)恒成立,则n的最大值?请给出详细的过程!谢谢!

Answer 1

1/(a-b)+1/(b-c)=(a-c)/(a-b)(b-c),
∴n≤(a-c)^2/(a-b)(b-c)恒成立,

∴只须求(a-c)^2/(a-b)(b-c)的最小值就可以了,即求(a-b)(b-c)的最大值
∵(a-b)>0,(b-c)>0,所以由均值不等式有：
(a-b)(b-c)≤{[(a-b)+(b-c)]/2}^2=(a-c)^2/4所以(a-b)(b-c)的最大值为(a-c)^2/4

∴(a-c)^2/(a-b)(b-c)最小值为4(a-c)^2/(a-c)^2=4,
即n的最大值为4

Answer 2

a>b>c，所以a-b>0，b-c>0，a-c>0
[1/(a-b)+1/(b-c)](a-c)
=[1/(a-b)+1/(b-c)][(a-b)+(b-c)]
=1+(a-b)/(b-c)+(b-c)/(a-b)+1
=2+[(a-b)/(b-c)+(b-c)/(a-b)]
≥2+2√[(a-b)/(b-c)*(b-c)/(a-b)]
=2+2
=4
即[1/(a-b)+1/(b-c)]≥4/(a-c)≥n/(a-c)恒成立
n≤4，n的最大值是4