已知x＞y,求证:x³－y³＞2x²y－2xy²

已知x＞y,求证:x³－y³＞2x²y－2xy²

证明：
x>y
x³-y³-(2x²y-2xy²)
=(x-y)(x²+xy+y²)-2xy(x-y)
=(x-y)(x²+xy+y²-2xy)
=(x-y)(x²-xy+y²)
=(x-y)[(x-y/2)²+3y²/4]
因为：x>y
所以：x-y>0
因为：(x-y/2)²+3y²/4>0
所以：x³-y³-(2x²y-2xy²)=(x-y)[(x-y/2)²+3y²/4]>0
所以：x³-y³>2x²y-2xy²