证明不等式已知x+y+z=1,x,y,z>0证明：x(1-2x)(1-3x)+y(1-2y)(1-3y)+z(1-2z)(1-3z)>=0

Question

证明不等式
已知x+y+z=1,x,y,z>0
证明：x(1-2x)(1-3x) + y(1-2y)(1-3y) +z(1-2z)(1-3z)>=0

Answer 1

已知x+y+z=1,x,y,z>0
证明：x(1-2x)(1-3x) + y(1-2y)(1-3y) +z(1-2z)(1-3z)>=0
x(1-2x)(1-3x)=x(y+z-x)(y+z-2x)
∑x(1-2x)(1-3x)
＝∑x(y+z-x)(y+z-2x)
＝∑x(yy+zz+2xx-3xy-3xz+2yz)
=2∑x^3+∑xyy+∑xzz-3∑xxy-3∑xxz+2∑xyz
=2(∑x^3-∑xyy-+∑xzz+3xyz)
=2∑x(x-y)(x-z)>=0
最后一式是舒尔（schur）不等式
舒尔(schur)不等式
已知x,y,z>=0
则
∑x(x-y)(x-z)>=0
舒尔(schur)不等式的证明：
不妨设x>=y>=z
∑x(x-y)(x-z)
=x(x-y)(x-z)+y(y-x)(y-z)+z(z-x)(z-y)
>=x(x-y)(x-z)+y(y-x)(y-z)
>=x(x-y)(y-z)+y(y-x)(y-z)
=(x-y)^2(y-z)
>=0
证毕!