- 设x+y+z=0,求证:6(x^3+y^3+z^3)=<(
- 求证:6(x^3+y^3+z^3)=<(x^2+y^2+z^2)^3.
- 证明:
由x+y+z=0及对称性,不妨设x>=0,y>=0,z=<0.
由x+y=-z得z^2=(x+y)^2,
从而(x^2+y^2+z^2)^3=8(x^2+xy+y^2)^3.
由A3>=G3,得
x^2+xy+y^2
=x(x+y)/2+y(x+y)/2+(x^2+y^2)/2
>=3[xy(x+y)^2/4*(x^2+y^2)/2]^(1/3)
>=3[(x^2y^2z^2)/4]^(1/3)
故(x^2+y^2+z^2)^3
>=3*27*(x^2y^2z^2)/4
=54x^2y^2z^2
=6(x^3+y^3+z^3)^2.
证毕.