Question

高中含参不等式问题，求助！在线等

已知a,b,c是正实数，abc=1
求证：(a-1+1/b)(b-1+1/c)(c-1+1/a) <= 1
尽量不要使用排序不等式，谢谢了

Answer 1

a=x/y,b=y/z,c=z/x,其中x,y,z都为正实数
即证
(x+y-z)(x+z-y)(z+y-x)<=xyz
(i)在(x+y-z),(x+z-y),(z+y-x)中若有一个<=0则其余两个必大于等于0则其积<=0<=xyz
(ii)在(x+y-z)(x+z-y)(z+y-x)全>0
则有
(x+y-z)(x+z-y)<=(x+y-z+x+z-y)^2/4
(即ab<=(a+b)^2/4)
可得(x+y-z)(x+z-y)<=x^2
同理还有两式
(x+y-z)(y+z-x)<=y^2
(y+z-x)(z+x-y)<=z^2
这三是相乘
[(x+y-z)(x+z-y)(y+z-x)]^2<=(xyz)^2
且(x+y-z)(x+z-y)(y+z-x)>0 xyz>0
则可得(x+y-z)(x+z-y)(y+z-x)<=xyz

(a-1+1/b)(b-1+1/c)(c-1+1/a) <= 1