Can we prove the inequality without opening the parentheses? $(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$ 
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)≥6(x^2+y^2+z^2)+3(xy+yz+xz)$$

I tried to use the following inequalities:
$$x^2+y^2+z^2≥xy+yz+xz$$
and The Cauchy–Schwarz inequality:
$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$
$$x^2+y^2+z^2≥x+y+z$$
I also tried
$$x+y+z ≥3\\ x^2+y^2+z^2≥3\\ xy+yz+xz≥3$$
But, I couldn't make progress.

I want to solve this inequality without expansion.  That is, without opening the parentheses.  Is this possible?


Increase the right hand side to $9(x^2+y^2+z^2)$ by switching $xy+yz+zx$ to $x^2+y^2+z^2$ (Cauchy).
Divide the left hand side by $xyz=1$.
This leads to $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$, which is easy to show. Namely, $x+y+z\geq 3\sqrt[3]{xyz}$ by AM-GM and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 3/\sqrt[3]{xyz}$ by HM-GM. This also shows that equality holds iff $x=y=z=1$.
