Let $x, y, z$ be positive distinct integers. Prove that $(x+y+z)(xy+yz+zx-2)\ge9xyz$

Question

I just came across the following question, in a book, which has as its topic contest-math:

Let $x, y, z$ be positive distinct integers. Prove that $(x+y+z)(xy+yz+zx-2)\ge9xyz$

I solved the question, in the following way:

Without loss of generality, from symmetry, we have that $x<y<z$.

However, $xy+yz+xz\ge\sqrt[3]{x^2y^2z^2}$ (AM-GM)

The equality holds if $x=y=z$. Impossible.

So, the following holds:

$xy+yz+xz\ge3*\sqrt[3]{x^2y^2z^2}+2$, since $y\ge x+1$ and $z\ge x+2$, so $y-1\ge x$ and $z-2\ge x$ and hence only this way can we have an $\ge$ in the inequality. (Which is false as proven in the comments)

We also have that $x+y+z\ge3*\sqrt[3]{xyz}$

So the inequality $(x+y+z)(xy+yz+xz-2)\ge3*\sqrt[3]{xyz}*3\sqrt[3]{x^2y^2z^2}=9xyz$, holds true.

I am not to certain about my proof and I also believe that there must exist far simpler solutions. Could you please verify that what I've written is indeed correct and show me some alternative methods?

Answer 1

Let $x<y<z$, $y=x+1+a$ and $z=x+2+a+b$,

where $a$ and $b$ are non-negative integers.

Thus, we need to prove that $$(x+y+z)(xy+xz+yz)-9xyz\geq2(x+y+z)$$ or $$\sum_{cyc}z(x-y)^2\geq2(x+y+z)$$ or $$(x+2+a+b)(1+a)^2+(x+1+a)(2+a+b)^2+x(b+1)^2\geq2(3x+3+2a+b).$$ Can you end it now?