Minimizing a function in $\mathbb R^3.$ 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.
 A: If $$f_1(x_1,y_1,z_1) = \frac{1}{2} \left[ (x_1-x_2)^2 + (x_2-x_3)^2+(x_3-x_1)^2\right],$$
There would be infinitely many solutions along the line
$$x_1=x_2=x_3.$$
But, alas, $ f(x_1, x_2, x_3) = f_1(x_1, x_2, x_3) +2x_2 x_3 =\frac{1}{2} \left[ (x_1-x_2)^2 + (x_2+x_3)^2+(x_3-x_1)^2\right].$
Now we have all non-negative terms, so the solution must be that all terms are zero.  Solving $x_1=x_2=-x_3=-x_1$, the global solution is $(0,0,0)$.
Had some help with this (see comments below).
A: Notice that, for each $u,v\in\mathbb{R}$, it is $uv\le \frac{1}{2}u^2+\frac{1}{2}v^2$, because $0\le(u-v)^2=u^2-2uv+v^2$. Hence, for each $u,v\in\mathbb{R}$ it is $-uv \ge -\frac{1}{2}u^2-\frac{1}{2}v^2$.
So, using this inequality three times, it is:
$$-x_1x_2-x_1x_3-x_2x_3\ge-\frac{1}{2}x_1^2-\frac{1}{2}x_2^2-\frac{1}{2}x_1^2-\frac{1}{2}x_3^2-\frac{1}{2}x_2^2-\frac{1}{2}x_3^2=-x_1^2-x_2^2-x_3^2$$
$$\implies x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3\ge0 \implies f(x_1,x_2,x_3)\ge0$$
For each $(x_1,x_2,x_3)\in\mathbb{R}^3$, and since $f(0,0,0)=0$ this means that $(0,0,0)$ is a global point of minimum in $\mathbb{R}^3$ for $f$.
