<s>[INST] Is this nice-looking inequality actually trivial? 
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz)≥2(x+y+z)^2  $$

I tried to use the inequality
$$x^2+y^2+z^2≥xy+yz+xz$$
Then I got,
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz) ≥2(xy+yz+xz)^2≥2(x+y+z)^2\implies xy+yz+xz≥x+y+z$$
Which is correct, I think. Because, the degree of the polynomial $xy+yz+xz$ is greater than the degree of the polynomial $x+y+z$. But, I am not sure.

Is this inequality trivial and is my solution correct?

 [/INST] Let $S=x+y+z, T=xy+xz+yz$ and note that $S,T \ge 3$ by AMGM and $xyz=1$ and also $S^2 \ge 3T$ by C-S
The inequality is: $T(S^2-T) \ge 2S^2$
Case 1: $S \ge T$ then $S^2(T-2) \ge S^2 \ge T^2$ which rewrites to the required inequality
Case2: $S \le T$ then using $S^2-T \ge 2T$ we get $T(S^2-T) \ge 2T^2 \ge 2S^2$ so we are done also!
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