Đáp án:
Bài 12:
$a)x^2+2xy+y^2+1=(x+y)^2+1≥1>0$
$b)x^2+y^2+1≥xy+x+y$
$↔2x^2+2y^2+2-2xy-2x-2y≥0$
$↔(x-y)^2+(y-1)^2+(x-1)^2≥0$ ( luôn đúng )
Dấu $"="$ xảy ra $↔x=y=1$
$c)x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac 14 + \dfrac 34=\bigg(x-\dfrac 12\bigg)^2+\dfrac 34≥\dfrac 34>0$