x3+y3-(x2y+xy2)
=x2(x-y)-y2(x-y)
=(x-y)(x2-y2)
=(x-y)(x+y)(x-y)
=(x-y)^2(x+y)
因为X,Y都是正实数
则x+y>0 (x-y)^2>=0
所以x3+y3-(x2y+xy2)>=0
x3+y3>=x2y+xy2
(x^3+y^3)/(x^2y+xy^2)=(x+y)(x^2-xy+y^2)/xy(x+y)=x/y + y/x -1 ≥2√(x/y*y/x)-1=2-1=1
∴x^3+y^3≥x^2y+xy^2
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