证明：对任意实数x,y,有x4+y4>=0.5xy（x+y）2

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2x^4+2y^4-xy(x+y)^2≥0
2x^4+2y^4≥xy(x+y)^2
x^4+y^4≥0.5xy(x+y)^2

1、x^4＋y^4≥2x²y² (x=y时取等号)；
2、x^4＋y^4≥x^3y＋xy^3。证明如下：
(x^4＋y^4)－(x^3y＋xy^3)=(x^4－x^3y)－(y^4－xy^3)=x^3(x－y)－y^3(x－y)=(x－y)(x^3－y^3)=(x－y)²(x²＋xy＋y²)=(x－y)²[(x＋1/2y)...