Not sure what to do after trying to simplify the inequality and getting nowhere? 
Here is the question:

If $x, y, z \in {\displaystyle \mathbb {R} }$, then prove
$$x^2+y^2+z^2+4\ge2(x+y+z)$$

Here is what I tried doing:

I tried simplifying the inequality and getting all the terms on one side, like so:
$$x^2+y^2+z^2+4\ge2x+2y+2z$$
$$x^2+y^2+z^2+4-2x+2y+2z\ge0$$
$$(x^2-2x)+(y^2-2y)+(z^2-2z)+4\ge0$$
$$(x^2-2x)+(y^2-2y)+(z^2-2z)\ge-4$$
Is it possible to use the fact that it is always true that $x^2-2x\ge-4$, $y^2-2y\ge-4$, and $z^2-2z\ge-4$, to show that inequality holds? Other than that, I'm not sure what to do next. I can't think of any manipulation that might work in this case.
Any help would be greatly appreciated!

\begin{align}
x^2 - 2x + y^2 - 2y + z^2 - 2z + 4
= (x-1)^2 + (y-1)^2 + (z-1)^2 + 1
\ge 1 > 0.
\end{align}
