Question

已知x,y都是正整数,求证x^3+y^3>=x^2y+xy^2

Answer 1

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