证明不等式(x2+y2)2>=xy(x+y)2

证明：(x^2+y^2)^2-xy(x+y)^2
=x^4+2x^2 y^2+y^4-xy(x^2+2xy+y^2)
=x^4-x^3 y+y^4-xy^3
=x^3(x-y)-y^3 (x-y)
=(x-y)^2 (x^2+xy+y^2)
=(x-y)^2 [(x+y/2)^2+3y^2/4]
显然在实数范围内上式≥0,故(x^2+y^2)^2≥xy(x+y)^2.