Help with Inequality Given that $x, y, z$ are nonnegative real numbers such that :
$$x^2 + y^2 + z^2 + xyz = 4$$
Prove that
$0 ≤ xy + yz + zx − xyz ≤ 2$

use this $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
and $$x^2+y^2+z^2+xyz=4$$
then we set
$$x=2\cos{A},y=2\cos{B},z=2\cos{C}$$
$$\Longleftrightarrow 0\le 4(\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{A}\cos{C})-8\cos{A}\cos{B}\cos{C}\le 2$$
and since
$$\cos{A}\cos{B}\cos{C}=\dfrac{s^2-(2R+r)^2}{4R^2},\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{C}\cos{A}=\dfrac{s^2+r^2-4R^2}{4R^2}$$
then 
$$\Longleftrightarrow 0\le\dfrac{s^2+r^2-4R^2}{R^2}-\dfrac{2}{R^2}\left(s^2-(2R+r)^2\right)\le 2$$
and left hand it suffices that
$$s^2+r^2-4R^2-2s^2+2(2R+r)^2\ge 0$$
$$\Longleftrightarrow 3r^2+8Rr+4R^2\ge s^2$$
use the Gerrseten inequality:
$$r(16R-5r)\le s^2\le 4R^2+4Rr+3r^2$$
then 
$$\Longleftrightarrow 4R^2+4Rr+3r^2\le 3r^2+4R^2+8Rr$$
$$\Longleftrightarrow 4Rr\ge 0$$is obvious.
and the Right Hand,
$$\Longleftrightarrow s^2+r^2-4R^2-2(s^2-(2R+r)^2)\le 2R^2 $$
$$\Longleftrightarrow 2R^2+8Rr+3r^2\le s^2$$
since
$$s^2\ge r(16R-5r)$$,
it follow that
$$r(16R-5r)\ge 2R^2+8Rr+3r^2$$
$$\Longleftrightarrow (R-2r)^2\ge 0$$
is obivous.
