Prove that $(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$ 
Let, $x,y,z>0$ such that $xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$$

My progress:
Using the Cauchy-Schwars inequality I got
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(xy+yz+xz)(x+y+z)≥2(x+y+z)^2 \implies xy+yz+xz≥x+y+z$$
But, this is not always true.
I also tried
$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$
I got
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥(xy+yz+xz)\left(\frac{(x+y+z)^2}{3}+x+y+z\right)≥2(x+y+z)^2 \implies (xy+yz+xz)(x+y+z+3)≥6(x+y+z)$$
But, again I failed.
 A: Let $S=x+y+z, T=xy+xz+yz$ and note that $S,T \ge 3$ by AMGM and $xyz=1$
The inequality is: $T(S^2+S-2T) \ge 2S^2$
Case 1: $S \ge T$ then $S^2(T-2) \ge S^2 \ge T^2 \ge T(2T-S)$  which rewrites to the required inequality
Case 2: $S \le T$; then we have $T(S^2+S-2T) \ge S(S^2+S-2T)$ so we need to prove only that $S^2+S-2T \ge 2S$ or that $S^2 \ge S+2T$. But (by C-S) $S^2 \le 3(S^2-2T)$ so $S^2 \ge 3T \ge 2T+S$ and we are done
A: pqr method:
Let $p = x + y + z, q = xy + yz + zx, r = xyz = 1$.
Using AM-GM, we have $p \ge 3\sqrt[3]{r} = 3$ and $q \ge 3\sqrt[3]{r^2} = 3$.
The inequality is written as
$$q(p^2 - 2q + p) \ge 2p^2$$
or
$$(q - 2)p^2 + q(-2q + p) \ge 0.$$
Using $q\ge 3$ and $p^2 \ge 3q$, it suffices to prove that
$$(q - 2)\cdot 3q + q(-2q + p) \ge 0$$
or
$$q(q + p - 6) \ge 0$$
which is true.
We are done.
A: This is not an answer! But this is what I have so far. One can consider symmetric polynomials $\sigma_1=x+y+z,\ \sigma_2=xy+yz+zz,\ \sigma_3=xyz=1\ $.
We see that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yx+xz)=\sigma_1^2-2\sigma_2.$ Using that we can rewrite your inequality in the following way.
$$\sigma_2(\sigma_1^2-2\sigma_2+\sigma_1)\geq2\sigma_1^2 \Leftrightarrow$$
$$\sigma_1^2\sigma_2-2\sigma_2^2+\sigma_1\sigma_2-2\sigma_1^2\geq0.$$
However, you can show that the following inequalities are true:  i) $\sigma_1^2\geq 3\sigma_2;$ 2) $\sigma_2^2\geq 3\sigma_2$; 3) $\sigma_1\sigma_2\geq 9$.
I believe that using these inequalties one can show that your inequality is also true.
