Question

急 已知a>b>c>0,求证;b/a-b>b/a-c>c/a-c

Answer 1

证明:
(a-c)^2=(a-b+b-c)^2=(a-b)^2+(b-c)^2+2*(a-b)(b-c)
>=2*(a-b)(b-c)+2*(a-b)(b-c)
=4*(a-b)(b-c)
即(a-c)^2>=4*(a-b)(b-c)
又a>b>c
所以(a-c)/(a-b)(b-c)>=4/(a-c)
即(a-b+b-c)/(a-b)(b-c)>=4/(a-c)
所以1/(a-b)+1/(b-c)>=4/(a-c)
所以1/(a-b)+1/(b-c)+4/(c-a)>=0

即1/(a-b)+1/(b-c)≥4/(a-

追问

没有看懂，能简单点吗
说明b/(a-b)>b/(a-c)就行，谢谢了

追答

设 a-b = x,b-c = y,则 a-c = x+y,其中 x,y 均为正实数。此时不等式化为(x+y)+1/(xy)≥ 4.容易看出(x+y)≥ 4xy,所以(x+y)+1/(xy)≥ 4xy+1/(xy)≥ 2根号[(4xy)*1/(xy)]= 4.等号成立当且仅当 x = y 且 4xy = 1/(xy),即 x = y = 根号2/2,即 a-b = b-c = ...