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.

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
$$g^4 f(g,g,z)=1+2g^5+g^2-4g^3=(g-1)^2(2g^3+4g^2+2g+1) \ge 0 \,.$$
