设x,y,z为正实数，x+y+z=1.求证: yz/x+zx/y+xy/z+9xyz>=1+x^2+y^2+z^2 (1) 证明 (1)式等价于 y^2*z^2+z^2*x^2+x^2*y^2+9(xyz)^2≥xyz+xyz(x^2+y^2+z^2) (2) 将(2)式齐次化处理得: (x+y+z)^2*(y^2*z^2+z^2*x^2+x^2*y^2)+9(xyz)^2≥ xyz(x+y+z)^3+xyz(x^2+y^2+z^2)*(x+y+z) (3) (3)展开化简为 Σx^4*(y^2+z^2)+2Σy^3*z^3-2xyzΣx^3-2xyzΣx^2*(y+z)+6(xyz)^2≥0 (4) 因为(4)式是全对称式,不失一般性,设x=max(x,y,z),(4)式分解为: 4y^2*z^2*(x-y)*(x-z)+[x^4+2x^3*(y+z)+x^2*(y^2+z^2)-2xyz(y+z)+y^2*z^2]*(y-z)^2≥0

设x,y,z为正实数，x+y+z=1.求证: yz/x+zx/y+xy/z+9xyz>=1+x^2+y^2+z^2 (1)证明 (1)式等价于y^2*z^2+z^2*x^2+x^2*y^2+9(xyz)^2≥xyz+xyz(x^2+y^2+z^2) (2)将(2)式齐次化处理得:(x+y+z)^2*(y^2*z^2+z^2*x^2+x^2*y^2)+9(xyz)^2≥xyz(x+y+z)^3+xyz(x^2+y^2+z^2)*(x+y+z) (3)(3)展开化简为Σx^4*(y^2+z^2)+2Σy^3*z^3-2xyzΣx^3-2xyzΣx^2*(y+z)+6(xyz)^2≥0 (4)因为(4)式是全对称式,不失一般性,设x=max(x,y,z),(4)式分解为:4y^2*z^2*(x-y)*(x-z)+[x^4+2x^3*(y+z)+x^2*(y^2+z^2)-2xyz(y+z)+y^2*z^2]*(y-z)^2≥0上式显然成立.证毕