<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] 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
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