已知x,y都是正整数,求证x^3+y^3>=x^2y+xy^2
问题描述：
已知x,y都是正整数,求证x^3+y^3>=x^2y+xy^2
答
X^3+Y^3=(X+Y)(X^2-XY+Y^2)
X^2Y+XY^2=(X+Y)XY
X^2+Y^2>=2XY (因（X-Y)^2>=0)
所以 X^2-XY+Y^2>=XY
(x,y都是正整数)
得(X+Y)(X^2-XY+Y^2)>=(X+Y)XY
即x^3+y^3>=x^2y+xy^2
答
x^3+y^3>=x^2y+xy^2
左边减去右边得：=x^2（x-y）+y^2（y-x）=(x-y)^2(x+y)
所以结果大于或者等于0
答
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
答
x^3+y^3-x^2y-xy^2=x^2(x-y)+y^2(y-x)=(x^2-y^2)(x-y)=(x-y)^2(x+y)>=0
所以x^3+y^3>=x^2y+xy^2