a>b>c,1/a-b+1/b-c≧m/a-c ,求m
问题描述：
a>b>c,1/a-b+1/b-c≧m/a-c ,求m
答
因 a>b>c
则 a-b>0,b-c>0,a-c>0
因 1/(a-b)+1/(b-c) ≥ m/(a-c)
(b-c+a-b)/(a-b)(b-c)≥ m/(a-c)
(a-c)^2/(a-b)(b-c)≥m
由基本不等式有
(a-b)(b-c)≤[[(a-b)+(b-c)]/2]^2 = (a-c)^2/4
则
(a-c)^2/(a-b)(b-c) ≥ 4
所以
m≤4