`(x + y + 1)^2 = 3(x^2 + y^2 + 1)`
`<=> 3(x^2 + y^2 + 1) - (x + y + 1)^2 = 0`
`<=> 3x^2 + 3y^2 + 3 - (x^2 + y^2 + 1 + 2xy + 2y + 2x)=0`
`<=> 3x^2 + 3y^2 + 3 - x^2 - y^2 - 1 - 2xy - 2y - 2x = 0`
`<=> 2x^2 + 2y^2 - 2xy - 2x - 2y + 2 = 0`
`<=> (x^2 - 2x + 1) + (y^2 - 2y + 1) + (x^2 - 2xy + y^2) = 0`
`<=> (x - 1)^2 + (y-1)^2 + (x - y)^2 = 0`
Ta có: $\begin{cases} (x - 1)^2 \ge 0 \ ∀ \ x\\ (y - 1)^2 \ge 0 \ ∀ \ y\\ (x-y)^2 \ge 0 \ ∀ \ x;y\end{cases}$
`=> (x - 1)^2 + (y - 1)^2 + (x-y)^2 ≥ 0`
Dấu $"="$ xảy ra `<=>` $\begin{cases} x-1=0\\y-1=0\\x-y=0\end{cases}$ `<=>` $\begin{cases} x=1\\y=1\end{cases}$
KL: .......
