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?

 A: 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!
A: Your solution is wrong, a value of a polynomial of smaller degree can be bigger than the corresponding value of a polynomial of bigger degree. For example $x+5$ has degree $1$, $x^9$ has degree $9$ but $1+5>1^9$.
Here is a correct solution. I would be glad to see a simpler solution, but I do not see a substantially different idea.
Denote $a=xy+yz+xz, b=x+y+z$. Your inequality is the same as $a(b^2-a)\ge 2b^2$ or $ab^2-a^2\ge 2b^2$ or $(a-2)b^2-a^2\ge 0$. Note that by $AM>GM, x^2+y^2+z^2\ge 3$, also $a\ge 3$.
Then $b^2=x^2+y^2+z^2+2a\ge 3+2a$. Hence your inequality will follow from $(a-2)(3+2a)-a^2\ge 0$ or $a^2-6-a\ge 0$. Since $a>0$, the latter inequality is equivalent to $a\ge 3$ which we have already established.
