$x^{2}$ + x + 1
= $x^{2}$ + 2 . $\dfrac{1}{2}$ . x + $(\dfrac{1}{2})^{2}$ - $(\dfrac{1}{2})^{2}$ + 1
= (x + $\dfrac{1}{2}$)² - $\dfrac{1}{4}$ + 1
= (x + $\dfrac{1}{2}$)² + $\dfrac{3}{4}$
Ta có: (x + $\dfrac{1}{2}$)² ≥ 0 ∀ x
⇔ (x + $\dfrac{1}{2}$)² + $\dfrac{3}{4}$ ≥ $\dfrac{3}{4}$ > 0 ∀ x
⇔ $x^{2}$ + x + 1 > 0 ∀ x
Vậy $x^{2}$ + x + 1 luôn dương với mọi x