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

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.

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

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

So, the following holds:

$$xy+yz+xz\ge3*\sqrt{x^2y^2z^2}+2$$, 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 $$x+y+z\ge3*\sqrt{xyz}$$

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

• yes sorry, thanks for picking me up on it
– user814992
Aug 11, 2020 at 16:07
• I've just edited it
– user814992
Aug 11, 2020 at 16:08
• $xy+yz+xz\ge3*\sqrt{x^2y^2z^2}+2$ isn't true. Your justification for the $+2$ is invalid, as shown by trying $(x,y,z)=(1,2,3)$ Aug 11, 2020 at 16:23
• Yes indeed you're right, I hadn't thought of that. Is there any way of finishing the question off the way I had started it?
– user814992
Aug 11, 2020 at 16:39

Let $$x, $$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?

• I'm trying to understand it but I can't understand why if we prove that \sum_{cyc}z(x-y)^2\geq2(x+y+z), then it is solved
– user814992
Aug 11, 2020 at 16:41
• Could you please explain it to me and also thanks a lot for the response
– user814992
Aug 11, 2020 at 16:42
• He just opened the bracket and simplified it Aug 11, 2020 at 16:42
• @Michael Christofer I think we can not end your way because AM-GM does not save the case of an equality occurring. Aug 11, 2020 at 16:47
• @Michael Christofer You are welcome! Good luck! Aug 11, 2020 at 16:49