Find the infimum and supremum of the set $G = \{x \in \Bbb R \mid x^2+x+1>0\}$.
Attempt: For all real $x$, the discriminant is $1^2-4(1)(1) =-3<0$. So, for all real $x$ has $x^2+x+1>0$ i.e. $G = \Bbb R$.
Hence, $G = (-\infty,\infty)$ and hence, $\inf G = -\infty$ and $\sup G = \infty$.
But, for another consideration, for all real $x$, we have $x^2+x+1 = (x+\frac{1}{2})^2 + \frac{3}{4} \ge \frac{3}{4}$, which equality attained when $x = - \frac{1}{2}$. Therefore, $\inf G = \frac{3}{4}$ and $\sup G = \infty$.
What should I do?
Seems you are mixing two different things here whose notations seem similar
(and maybe that's what confused you) but they are not the same thing.
$G_1 = \{x \in \Bbb R \mid x^2+x+1>0\}$
$G_2 = \{x^2+x+1 \mid x \in \Bbb R \}$
Note that $$G_1 = \Bbb R$$ while
$$G_2 = [3/4; \infty)$$
So $$\inf G_2 = 3/4$$
and so there is no contradiction here or any logical issue.