<s>[INST] 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.
 [/INST] 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.
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