Đặt : `A=x^2+xy+y^2-3x-3y`
`⇒A=x^2+xy+y^2-2x-x-2y-y`
`⇒A=(x^2-2x+1)+(y^2-2y+1)+(xy-x-y+1)-3`
`⇒A=(x-1)^2+(y-1)^2+[x(y-1)-(y-1)]-3`
`⇒A=(x-1)^2+(y-1)^2+(x-1)(y-1)-3`
`⇒A=(x-1)^2+1/4(y-1)^2+3/4(y-1)^2+2 .1/2(x-1)(y-1)-3`
`⇒A=[(x-1)^2+2.1/2(x-1)(y-1)+1/4(y-1)^2]+3/4(y-1)^2-3`
`⇒A=[(x-1)+1/2(y-1)]^2+3/4(y-1)^2-3≥-3 ∀x`
Dấu $"="$ xảy ra khi :
$\begin{cases}[(x-1)+\dfrac12(y-1)]^2=0\\\dfrac34(y-1)^2=0\end{cases}$
$⇒\begin{cases}(x-1)+\dfrac12(y-1)=0\\y-1=0\end{cases}$
`⇒x=-1`
