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.
 A: Let $$f(x,y,z):=x^2 + y^2 + z^2 + x + y + z - 2(xy+yz+zx) \,.$$
Suppose WLOG that $x \ge y \ge z$, and write $g=\sqrt{xy}\,. \;$
Then
\begin{eqnarray} f(x,y,z)-f(g,g,z)&=& x^2+y^2-2g^2+(x+y-2g)(1-2z) \\ &=&
(\sqrt{x}-\sqrt{y})^2 \cdot \Bigl( (\sqrt{x}+\sqrt{y})^2+1-2z\Bigr) \ge 0 \,,
\end{eqnarray}
since $\sqrt{x}+\sqrt{y} \ge 2\sqrt{z}. \;$ Moreover,
$$f(g,g,z)=z^2+2g+z-4gz \,,$$
so, recalling that $zg^2=zxy=1$, we get
$$g^4 f(g,g,z)=1+2g^5+g^2-4g^3=(g-1)^2(2g^3+4g^2+2g+1) \ge 0 \,.$$
A: Let $x=a^3, y=b^3, z=c^3$ then we have that $abc=1$ and we want to prove that $a^6+b^6+c^6+(a^3+b^3+c^3)=a^6+b^6+c^6+abc(a^3+b^3+c^3)\ge 2(a^3b^3+b^3c^3+c^3a^3)$. Now by Schur's inequality for $t=4$, $a^6+b^6+c^6+abc(a^3+b^3+c^3)\ge \sum_{sym} a^5b$ and by Muirhead's inequality $\sum_{sym} a^5b \ge \sum_{sym} a^3b^3=2(a^3b^3+b^3c^3+c^3a^3)$ and thus we are done.
A: 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 $p^3 - 4pq + 9r \ge 0$.
