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