Question

证明：对任意实数x,y,有x4+y4>=0.5xy（x+y）2

Answer 1

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