题目

已知x,y都是正整数,求证x^3+y^3>=x^2y+xy^2

提问时间：2021-01-13

答案

x^3+y^3-x^2y-xy^2
=(x^3-x^2y)+(y^3-xy^2)
=x^2(x-y)+y^2(y-x)
=x^2(x-y)-y^2(x-y)
=(x-y)(x^2-y^2)
=(x-y)^2(x+y) x,y都是正整数,
>=0
所以x^3+y^3>=x^2y+xy^2