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

Let $x=\frac{2a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{2b}{\sqrt{(a+b)(b+c)}}$, where $a$, $b$ and $c$ be positives.
Hence, $z=\frac{2c}{\sqrt{(a+c)(b+c)}}$ and the left inequality it's
$$4\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}\geq\frac{8abc}{(a+b)(a+c)(b+c)}$$ or
$$\sum_{cyc}ab\sqrt{(a+c)(b+c)}\geq2abc,$$
which is true because
$$\sum_{cyc}ab\sqrt{(a+c)(b+c)}\geq\sum_{cyc}abc\geq2abc.$$
The right inequality.
We need to prove that
$$4\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}-\frac{8abc}{(a+b)(a+c)(b+c)}\leq2$$ or
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq(a+b)(a+c)(b+c)+4abc,$$
which is AM-GM:
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}ab(a+c+b+c)=$$
$$=\sum_{cyc}(a^2b+a^2c+2abc)=(a+b)(a+c)(b+c)+4abc.$$
Done!
