Question

设a＞b＞c，且1/(a-b)＋1/(b-c)＞m/(a-c)恒成立，求m的取值范围

Answer 1

1/(a-b)+1/(b-c)≥m/(a-c) （两边同时乘以a-c）
<==>(a-c)/(a-b)+(a-c)/(b-c)≥m
<==>只需求得左边的取值范围（或最小值）即可
左边=(a-b+b-c)/(a-b)+(a-b+b-c)/(b-c)
=1+(b-c)/(a-b)+(a-b)/(b-c)+1
=2+(b-c)/(a-b)+(a-b)/(b-c)
≥2+2sqrt[(b-c)/(a-b)]*[(a-b)/(b-c)]（sqrt是开平方,用到均值不等式）
=4
所以4≥m