<s>[INST] 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?

 [/INST] Rewrite the left hand side as follows:
$$
(x+y+z)(xy+yz+xz)(x^2+y^2+z^2)=
\\
=\left(\frac{2}{3}(x+y+z)(xy+yz+zx)\right)\cdot(x^2+y^2+z^2)+
\\
+\left(\frac{1}{3}(x+y+z)(x^2+y^2+z^2)\right)\cdot(xy+yz+zx).
$$
Can you continue now?
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