设x,y,z∈R,2x+2y+z+8=0,则(x-1)^2+(y+2)^2+(z-3)^2最小值

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设x,y,z∈R,2x+2y+z+8=0,则(x-1)^2+(y+2)^2+(z-3)^2最小值

设x,y,z∈R,2x+2y+z+8=0,则(x-1)^2+(y+2)^2+(z-3)^2最小值
设x,y,z∈R,2x+2y+z+8=0,则(x-1)^2+(y+2)^2+(z-3)^2最小值

设x,y,z∈R,2x+2y+z+8=0,则(x-1)^2+(y+2)^2+(z-3)^2最小值
由柯西不等式知:
[(x-1)²+(y+2)²+(z-3)²](2²+2²+1²)
≥[2(x-1)+2(y+2)+(z-3)]²
=(2x+2y+z-1)²
=(-8-1)²=81
∴(x-1)²+(y+2)²+(z-3)²≥81/(2²+2²+1²)=9
∴最小值为9