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 $x=a^3, y=b^3, z=c^3$ 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.
