I don't know why you need multiple proofs for a simple result, but here are two overkilling solutions. The first one utilizes some knowledge in linear algebra. The second one uses Euclidean geometry along with some trigonometry.
Consider the matrix $\mathbf{B}:=\begin{bmatrix}1&\frac12\\\frac12&1\end{bmatrix}$. Being a real symmetric $2$-by-$2$ matrix, $\mathbf{B}$ has two real eigenvalues, which are $\frac{3}{2}$ and $\frac{1}{2}$. As both eigenvalues are positive, $\mathbf{B}$ is a positive-definite matrix, whence it induces a positive-definite symmetric bilinear form $\langle\_,\_\rangle:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ sending a pair $(\mathbf{u},\mathbf{v})$ of $2$-by-$1$ column vectors in $\mathbb{R}^2$ to
$$\langle \mathbf{u},\mathbf{v}\rangle:=\mathbf{u}^\top\,\mathbf{B}\,\mathbf{v}\,.$$
That is, $$\langle \mathbf{u},\mathbf{u}\rangle \geq 0\text{ for all }\mathbf{u}\in\mathbb{R}^2\,,$$
and the inequality becomes an equality iff $\mathbf{u}$ is the zero vector. In particular, when $\mathbf{u}=(x,1)$, where $x$ is an arbitrary real number, we get $\mathbf{u}\neq \boldsymbol{0}$, whence
$$x^2+x+1=\langle\mathbf{u},\mathbf{u}\rangle>0\,,$$
as desired.
Alternatively, consider three points in $\mathbb{R}^2$: the origin $O=(0,0)$, the point $A=(1,0)$, and the point $B=\left(-\frac{x}{2},\frac{\sqrt{3}x}{2}\right)$. Note that $\angle AOB=\frac{2\pi}{3}$ for $x>0$, and $\angle {AOB}=\frac{\pi}{3}$ for $x<0$. Using the Law of Cosine, you get $$AB^2=x^2+x+1\,,$$
whence $x^2+x+1>0$, noting that $A\neq B$ for any value of $x$. (The case $x=0$ can be checked separately, but then $x^2+x+1=AB^2=1>0$ still holds.)