`F = x^2 + y^2 - 8x + 2y + 20`
`= x^2 - 8x + 16 + y^2 + 2y + 1 + 3`
`= (x - 4)^2 + (y + 1)^2 + 3`
Do `{((x - 4)^2 >= 0),((y + 1)^2 >= 0):}`
`=> (x - 4)^2 + (y + 1)^2 >= 0`
`=> (x - 4)^2 + (y + 1)^2 + 3 >= 3`
Hay `F >= 3`
Dấu = có khi:
`{(x - 4= 0),(y + 1= 0):}`
`<=> {(x =4),(y =-1):}`
Vậy ...