把右边的式子移到左边,并乘以2倍得到：
x^4+y^4-0.5xy×（x+y）^2≥0证明这个不等式成立就可以了.
因为 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




^,符号代表方的意思,所以就是这样喽!