I am currently solving the following task:
Exercise. Minimize in $\mathbb R^3$ the function $f(x_1,x_2,x_3) = x_1^2 + x_2^2 + x_3^2 - x_1x_2 + x_2x_3 - x_1x_3.$
My attempt. I was able to show that $(0,0,0)$ is a local strict minimizer (consequently, $f(0,0,0) = 0$ is a local strict minimum) using the usual arguments, i.e., I found the critical points of $f$ (which is only one, $(0,0,0)$); then, I found the hessian of $f$ in the point $(0,0,0)$ and proved it was positive defined. Therefore, from the theorem below I am able to guarantee that $(0,0,0)$ is a local strict minimizer.
Theorem. Assume that $f$ is twice differentiable in $\bar x \in X.$ Then, if $\nabla f(\bar x) = 0$ and $H(\bar x) = \nabla ^2 f(\bar x)$ is positive defined, then $\bar x$ is a local strict minimizer.
My question. After this, one question comes up to my mind: is $(0,0,0)$ a global minimizer? Looking to answer this further question, I gave some values to some of the coordinates ($x_1, x_2 $ or $x_3$) individually and plotted the resultant graphs (which are $3D$). Every one I plotted (and I plotted a nice quantity...) leads me to the same conclusion: $(0,0,0)$ is, in fact, a global minimizer. After this, I tried to prove analitically that $f(x_1,x_2,x_3) \geqslant 0,$ for every $(x_1,x_2,x_3) \in \mathbb R^3$ (to show that, in fact $(0,0,0)$ is a global minimizer). But I am having a real hard time doing this. Any help towards proving this is apreciatted in advance.