问题如下：
已知1<a<=b<=c,
求证:
log(b)/log(a)+log(c)/log(b)+log(a)/log(c)<=log(a)/log(b)+log(b)/log(c)+log(c)/log(a)
第1个回答 推荐于2016-08-18
考虑1<a<=b<=c 则loga,logb,logc 可以为任意大於1的数
改写题目,设loga=x logb=y logc=z 1<x<=y<=z
y/x+z/y+x/z<= x/y+y/z+z/x
y+x<=z+y
(z-y)(y+x)(y-x)<=(z-y)(z+y)(y-x)
(z-y)(y^2-x^2)<=(z^2-y^2)(y-x)
z(y^2-x^2)-y(y^2-x^2)<=y(z^2-y^2)-x(z^2-y^2)
z(y^2-x^2)+x(z^2-y^2)<=y(z^2-y^2)+y(y^2-x^2)
z(y^2-x^2)+x(z^2-y^2)<=y(z^2-x^2)
两边除xyz
(y^2-x^2)/xy+(z^2-y^2)/zy<=(z^2-x^2)/xz
(y/x-x/y=(y^2-x^2)/xy,其馀同理)
y/x-x/y+z/y-y/z<=z/x-x/z
移项得证
y/x+z/y+x/z<=x/y+y/z+z/x追问
改写题目,设loga=x logb=y logc=z 1<x<=y<=z
y/x+z/y+x/z<= x/y+y/z+z/x
y+x<=z+y
(z-y)(y+x)(y-x)<=(z-y)(z+y)(y-x)
(z-y)(y^2-x^2)<=(z^2-y^2)(y-x)
z(y^2-x^2)-y(y^2-x^2)<=y(z^2-y^2)-x(z^2-y^2)
z(y^2-x^2)+x(z^2-y^2)<=y(z^2-y^2)+y(y^2-x^2)
z(y^2-x^2)+x(z^2-y^2)<=y(z^2-x^2)
两边除xyz
(y^2-x^2)/xy+(z^2-y^2)/zy<=(z^2-x^2)/xz
(y/x-x/y=(y^2-x^2)/xy,其馀同理)
y/x-x/y+z/y-y/z<=z/x-x/z
移项得证
y/x+z/y+x/z<=x/y+y/z+z/x追问
谢谢
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