Đáp án:
$a)$
$x^2+2xy+y^2+1=(x+y)^2+1≥1>0$
$b)$
$x^2+y^2+1≥xy+x+y$
$↔2x^2+2y^2+2≥2xy+2x+2y$
$↔(x^2+y^2-2xy)+(x^2-2x+1)+(y^2-2y+1)≥0$
$↔(x-y)^2+(x-1)^2+(y-1)^2≥0$
Dấu $"="$ xảy ra $↔x=y=1$
$c)$
$x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\bigg(x-\dfrac{1}{2}\bigg)^2+\dfrac{3}{4}≥\dfrac{3}{4}>0$