证明不等式(x2+y2)2>=xy(x+y)22都代表平方

原式=(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。

(x^2+y^2)^2-xy(x+y)^2=x^4-x^3y+y^4-xy^3=(x-y)(x^3-y^3)=(x-y)^2*(x^2+xy+y^2)=(x-y)^2*((x-y/2)^2+3/4*y^2)>=0,得证

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.

【(x^2+y^2)^2】- 【xy(x+y)^2】
= (x^2+y^2)^2- xy(x^2+y^2+2xy)
= (x^2+y^2)^2- xy(x^2+y^2+2xy)
= (x^2+y^2)^2- xy(x^2+y^2) - 2x^2y^2
= (x^2+y^2)(x^2+y^2-xy) - 2x^2y^2
= (x^2+y^2){(x^2+y^2-2xy)+xy} - 2x^2y^2
= (x^2+y^2){(x-y)^2+xy) - 2x^2y^2
= (x^2+y^2)(x-y)^2 + xy(x^2+y^2) - 2x^2y^2
= (x^2+y^2)(x-y)^2 + xy(x^2+y^2 - 2xy)
= (x^2+y^2)(x-y)^2 + xy(x-y)^2
= (x-y)^2 (x^2+y^2 + xy)
= (x-y)^2 {(x+1/2 y)^2 +3/4 y^2 }
∵ (x-y)^2≥0，(x+1/2 y)^2≥0，3/4 y^2≥0
∴ (x-y)^2 {(x+1/2 y)^2 +3/4 y^2 } ≥0
∴【(x^2+y^2)^2】- 【xy(x+y)^2】≥ 0
∴(x^2+y^2)^2 ≥ xy(x+y)^2