题目

已知x,y∈R+,且x≠y,求证x^5y^-5+x^-5y^5＞x^4y^-4+x^-4y^4
大家帮帮忙!

提问时间：2021-01-02

答案

作差法：
(x/y)^5+(y/x)^5-(x/y)^4-(y/x)^4
=(x^10+y^10-x^9y-xy^9)/(x^5y^5)
=(x^9-y^9)(x-y)/(x^5y^5)
=(x^3-y^3)(x^6+x^3y^3+y^6)(x-y)/(x^5y^5)
=(x-y)(x^2+xy+y^2)(x^6+x^3y^3+y^6)(x-y)/(x^5y^5)
=(x-y)^2(x^2+xy+y^2)(x^6+x^3y^3+y^6)/(x^5y^5)
因为x,y∈R+,且x≠y
所以(x-y)^2>0,(x^2+xy+y^2)>0,(x^6+x^3y^3+y^6)>0,(x^5y^5)>0
所以(x-y)^2(x^2+xy+y^2)(x^6+x^3y^3+y^6)/(x^5y^5)>0
所以(x/y)^5+(y/x)^5>(x/y)^4+(y/x)^4