Question

设x,y,z∈R，且x+y+z=1，求(x-1)²+(y+1)²+(z+1)²的最小值

Answer 1

x+y+z=1,
所以z=1-x-y,代入(x-1)²+(y+1)²+(z+1)²得
(x-1)^2+(y+1)^2+(2-x-y)^2
=(x-1)^2+y^2+2y+1+(2-x)^2-(4-2x)y+y^2
=2y^2+2(x-1)y+(x-1)^2+1+(2-x)^2
=2[y+(x-1)/2]^2+(x-1)^2/2+(x-2)^2+1
>=(1/2)[x^2-2x+1+2(x^2-4x+4)+2]
=(1/2)(3x^2-10x+11)
=(3/2)(x-5/3)^2+4/3,
所以当x=5/3,y=-(x-1)/2=-1/3,z=-1/3时它取最小值4/3.
解2利用柯西不等式
(x-1)²+(y+1)²+(z+1)²>=(1/3)(x-1+y+1+z+1)^2=4/3,
当x-1=y+1=z+1，x+y+z=1，即x=5/3,y=z=-1/3时取等号，
所以它的最小值是4/3.