证明：
1）因 x^5+y^5=2,容易证明,x+y>0 ①
2）(x+y)^5
=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5
=x^5+y^5+5(x^4y+xy^4)+10(x^3y^2+x^2y^3) ②
而 x^5+y^5-x^4y-xy^4
=(x-y)(x^4-y^4)
=(x^2+y^2)(x+y)(x-y)^2>=0 (x+y>0)
所以 x^5+y^5>x^4y^-xy^4
又 x^5+y^5-x^3y^2-x^2y^3
=(x+y)(x-y)^2*(x^2+xy+y^2)>=0
则由②式得
(x+y)^5=x^5+y^5+5(x^4y+xy^4)+10(x^3y^2+x^2y^3)