Question

如果正实数x、y、z满足x^3+y^3+z^3-3xyz=1,求x^2+y^2+z^2的最小值

"正实数"改为"非负实数"

Answer 1

由对烂激局称性， 无妨设x>=y>=z>=0; 显然不能全部相等 ，否则0=1 ，不可能；
条件化为 (x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]=2,
即：[(x-y)^2+(y-z)^2+(z-x)^2]=2/(x+y+z)
所以 x^2+y^2+z^2={(x+y+z)^2+ [(x-y)^2+(y-z)^2+(z-x)^2]}/饥让3
={(x+y+z)^2+2/(x+y+z)}/3={(x+y+z)^2+1/(x+Y+z)+1/(x+Y+z)}/3
>={(x+y+z)^2*1/(x+Y+z)*1/(x+Y+z)}^(1/3)=1
取等号条件是x+y+z=1, x^2+z^2+z^2=1, xy+xz+yz=0,从而铅尺 x(1-x)+yz=0,
所以 y=z=0,x=1 ，时最小值为1