Proving $x^2 + y^2 + z^2 + x + y + z \geq 2(xy+yz+zx)$ for positive values with $xyz=1$ 
How can you prove that
$$x^2 + y^2 + z^2 + x + y + z \geq 2(xy+yz+zx)$$
given that $x,y,z > 0$ and $xyz = 1$.

We can easily prove that the equality holds when $x = y = z = 1$
I could able to prove the result when one of $x,y,z$ is $1$ considering $x,y,z$ as $x,1/x, 1$ using the inequality $x + 1/x \geq 2$ for any positive number of $x$.
But couldn't able to find a full proof.

pqr method:
Let $p = x + y + z, \, q = xy + yz + zx, \, r = xyz = 1$.
It suffices to prove that $p^2 - 2q + p \ge 2q$ or
$$p^2 + p - 4q \ge 0.$$
Using $p^3 - 4pq + 9r \ge 0$ (three degree Schur), we have
$$q \le \frac{p^3 + 9}{4p}.$$
It suffices to prove that
$$p^2 + p - 4\cdot \frac{p^3 + 9}{4p} \ge 0$$
or
$$(p - 3)(p + 3)/p \ge 0$$
which is true since $p\ge 3\sqrt[3]{r} = 3$ (AM-GM).
We are done.
Remark: Three degree Schur inequality is
$$x(x - y)(x - z) + y(y - z)(y - x) + z(z - x)(z - y) \ge 0.$$
In pqr language (or substitution), it is $p^3 - 4pq + 9r \ge 0$.
