1、证明x³+y³≥x²y+y²x

1)x³+y³-(x²y+y²x)=x^2(x-y)+y^2(y-x)=(x-y)(x^2-y^2)=(x+y)(x-y)^2
原式在xy为正数时成立,你把条件漏了.
2)a³-b³=a²-b² =>a^2+ab+b^2=a+b =>(a+b)^2-ab=a+b
=>(a+b)^2-(a+b)=ab>0 =>a+b>1
同时(a+b)^2-(a+b)=ab3(a+b)^2/4-(a+b)a+b