x^4+y^4-(x³y+xy³)
=x³(x-y)-y²(x-y)
=(x-y)(x³-y³)
=(x-y)²(x²+xy+y²)
=(x-y)²((x+y/2)²+3/4*y²)
因为x≠y
所以(x-y)²>0,(x+y/2)²+3/4*y²>0
所以x^4+y^4-(x³y+xy³)>0
所以x^4+y^4>x³y+xy³