Question

已知x＞y,求证x³-y³＞2x²y-2xy²

Answer 1

x³-y³-(2x²y-2xy²)=x³-y³-2xy(x-y)=(x-y)(x^2+xy+y^2)-2xy(x-y)

=(x-y)(x^2-xy+y^2)

x>y>=0,(x-y)(x^2-xy+y^2)=(x-y)[x(x-y)+y^2)]>0

x>=0>y,(x-y)(x^2-xy+y^2)=(x-y)[x^2+x(-y)+y^2)]>0

0>=x>y,(x-y)(x^2-xy+y^2)=(x-y)[(x-y)^2+xy]>0

x＞y,所以x³-y³＞2x²y-2xy²