<s>[INST] How prove this $x^2+y^2+z^2+3\ge 2(xy+yz+xz)$ let $x,y,z$ be real numbers, and such $$x+y+z+xyz=4$$ show that
$$x^2+y^2+z^2+3\ge 2(xy+yz+xz)$$
My try: let
$$x+y+z=p,xy+yz+xz=q,xyz=r$$
then
$$p+r=4$$
then
$$\Longleftrightarrow p^2+3\ge 4q$$
But I can't.Thank you
 [/INST] Note that it is sufficient to consider $x, y, z \ge 0$.  Then the condition yields $4 = x+y+z+xyz \ge 3\sqrt[3]{xyz}+xyz \implies xyz \le 1$.
Also note that among $x-1, y-1, z-1$, at least two have the same sign.  WLOG let $(y-1)(z-1) \ge 0$.  Then we have:
$$(x-1)^2+(y-z)^2+2x(y-1)(z-1) \ge 0$$
$$\implies x^2+y^2+z^2+2xyz + 1 \ge  2(xy+yz+zx)$$
which along with $xyz \le 1$ gives you the result.
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