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

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

人气：254 ℃　时间：2020-05-04 13:33:19

解答

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+1...可否再帮我求出此时的xyz值当且仅当(x-1)/2=(y+2)/2=(z-3)/1，即x=2z-5,y=2z-8∴2x+2y+z+8=4z-10+4z-16+z+8=9z-18=0∴z=2,x=2z-5=-1,y=2z-8=-4