Does $1 + \frac{1}{x} + \frac{1}{x^2}$ have a global minimum, where $x \in \mathbb{R}$? Does the function 
$$f(x) = 1 + \frac{1}{x} + \frac{1}{x^2},$$
where $x \in {\mathbb{R} \setminus \{0\}}$, have a global minimum?
I tried asking WolframAlpha, but it appears to give an inconsistent result.

$g(x)=1+\frac{1}{x}+\frac{1}{x^2},x\ne 0$
$g'(x)=-\frac{1}{x^2}-2\frac{1}{x^3}=0\Rightarrow \frac{-2}{x^3}=\frac{1}{x^2}\Rightarrow x=-2$
$g''(x)=\frac{2}{x^3}+\frac{6}{x^4}|_{x=-2}> 0$
$\Rightarrow g(-2)$ is the minimum.
Edit:This shows that $g$ has a local minimum at $x=-2$ but this is also the global minimum because 
$g(x)-g(-2)=1+\frac{1}{x}+\frac{1}{x^2}-1-(\frac{-1}{2})-\frac{1}{2^2}=\frac{x+2}{2x}+\frac{(2-x)(2+x)}{4x^2}=(x+2)\frac{2x+2-x}{4x^2}=\frac{(x+2)^2}{4x^2}\ge0$
so $g(x)\ge g(-2)$ so $g(-2)$ is a global minimum.
