设x+y+z=1,则x2+xy+y2+y2+yz+z2+ z2+zx+x2的最小值为(  )

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/08 10:09:45
设x+y+z=1,则x2+xy+y2+y2+yz+z2+ z2+zx+x2的最小值为(  )

设x+y+z=1,则x2+xy+y2+y2+yz+z2+ z2+zx+x2的最小值为(  )
设x+y+z=1,则x2+xy+y2+y2+yz+z2+ z2+zx+x2的最小值为(  )

设x+y+z=1,则x2+xy+y2+y2+yz+z2+ z2+zx+x2的最小值为(  )
由x+y+z=1,两边平方,得:x^2+y^2+z^2+2xy+2xz+2yz=1.
所以:x^2+xy+y^2+Y^2+yz+z^2+z^2+zx+x^2
=(x^2+y^2+z^2+2xy+2xz+2yz)+(x^2+y^2+z^2-xy-xz-yz)
=1+[(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)]/2
=1+[(x-y)^2+(y-z)^2+(x-z)^2]/2
≥1
即:x^2+xy+y^2+Y^2+yz+z^2+z^2+zx+x^2的最小值是1.

1

x+y+z=1
(x+y+z)^2=1=x^2+y^2+z^2+2xy+2xz+2yz
x2+xy+y2+y2+yz+z2+ z2+zx+x2
=x^2+y^2+z^2+2xy+2xz+2yz+(x^2+y^2+z^2-xy-xz-yz)
=1+(x^2+y^2+z^2-xy-xz-yz)
=1+1/2(2x^2+2y^2+2z^2-2xy-2xz-2yz)
=1+1/2[(x-y)^2+(y-z)^2+(z-x)^2]≥3/2
当且仅当x=y=z=1/3等号成立