设a>b>c,1/(a-c)+1/(b-c)>=m/(a-c)恒成立,求m的取值范围?急用!麻烦您给做一下!结果为m

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/09 07:30:58
设a>b>c,1/(a-c)+1/(b-c)>=m/(a-c)恒成立,求m的取值范围?急用!麻烦您给做一下!结果为m

设a>b>c,1/(a-c)+1/(b-c)>=m/(a-c)恒成立,求m的取值范围?急用!麻烦您给做一下!结果为m
设a>b>c,1/(a-c)+1/(b-c)>=m/(a-c)恒成立,求m的取值范围?急用!麻烦您给做一下!
结果为m

设a>b>c,1/(a-c)+1/(b-c)>=m/(a-c)恒成立,求m的取值范围?急用!麻烦您给做一下!结果为m
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+2根号[(b-c)/(a-b)]*[(a-b)/(b-c)](均值不等式)
=4
所以4≥m

1/(a-c)+1/(b-c)>=m/(a-c)
1/(b-c)>=(m-1)/(a-c)
(a-c)>=mb-mc-b+c
(a-2c+b)/b>=m