你是題目應(yīng)是x^4+y^4≥0.5xy×（x+y）^2吧
如果是的話,把右邊的式子移到左邊,并乘以2倍得到：
x^4+y^4-0.5xy×（x+y）^2≥0證明這個不等式成立就可以了.
因為 2x^4+2y^4-xy×（x+y）^2
= 2x^4+2y^4-xy(x^2+2xy+y^2)
= 2x^4+2y^4-x^3y-2(xy)^2-xy^3
= x^4+y^4-2(xy)^2+x^4+y^4-x^3y-xy^3
= (x^2-y^2)^2+x^3(x-y)-y^3(x-y)
= (x^2-y^2)^2+(x-y)(x^3-y^3)
= (x^2-y^2)^2+(x-y)^2(x^2+xy+y^2)≥0
所以原不等式成立