<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] 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.
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