`Gth => (a-1)(b-1)(c-1)=4`

KMTTQ giả sử `a>=b>=c` mà `(a-1)(b-1)(c-1) >0`

`=>a>1`

`a^2+b^2+c^2=a^2-2+(b^2+1)+(c^2+1)`

`>= a^2-2+(b-1)^2/2+(c-1)^2/2`

`=a^2-2+((b-1)^2+(c-1)^2)/2`

`>=a^2-2+(b-1)(c-1)`

`=(a-2)^2+4(a-1)+(b-1)(c-1)-2`

`>=4(a-1)+(b-1)(c-1)-2`

`>=2 sqrt{4(a-1)(b-1)(c-1)}-2`

`=2 sqrt 16-2`

`=8-2=6`

Dấu `=` xảy ra `<=> a=2,b=c=-1`