Cách giải:
$A=x^2+y^2-xy-x+y+1$
$→4A=4x^2+4y^2-4xy-4x+4y+4$
$→4A=(4x^2-4xy+y^2)+3y^2-4x+4y+4$
$→4A=(2x-y)^2-2(2x-y)+3y^2+2y+4$
$→4A=(2x-y)^2-2(2x-y)+1+3(y^2+\dfrac{2}{3}y)+3$
$→4A=(2x-y-1)^2+3(y^2+2.y.\dfrac{1}{3}+\dfrac{1}{9})-\dfrac{1}{3}+3$
$→4A=(2x-y-1)^2+3(y-\dfrac{1}{3})^2+\dfrac{8}{3}$
$(2x-y-1)^2 \geq 0$
$3(y-\dfrac{1}{3})^2 \geq 0$
$→(2x-y-1)^2+3(y-\dfrac{1}{3})^2 \geq 0$
$→(2x-y-1)^2+3(y-\dfrac{1}{3})^2+\dfrac{8}{3} \geq \dfrac{8}{3}$
$→4A \geq \dfrac{8}{3}$
$→A \geq \dfrac{2}{3}$
Dấu = xảy ra khi
$\begin{cases}2x-y-1=0\\y=\dfrac{1}{3}\\\end{cases}$
$→\begin{cases}x=\dfrac{y+1}{2}=\dfrac{2}{3}\\y=\dfrac{1}{3}\\\end{cases}$
Vậy $GTNN_A=\dfrac{2}{3}↔\begin{cases}x=\dfrac{2}{3}\\y=\dfrac{1}{3}\\\end{cases}$
