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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!

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    $\begingroup$ $x^2-2x\ge-1$, so $(x^2-2x)+(y^2-2y)+(z^2-2z)\ge-3$ $\endgroup$ – J. W. Tanner Jan 14 '20 at 1:45
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\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}

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If you borrow three ones, you get $$ (x^2-2x)+(y^2-2y)+(z^2-2z)+4=(x-1)^2+(y-1)^2+(z-1)^2+1\geq1. $$

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