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?

  • $\begingroup$ yes sorry, thanks for picking me up on it $\endgroup$
    – user814992
    Aug 11, 2020 at 16:07
  • $\begingroup$ I've just edited it $\endgroup$
    – user814992
    Aug 11, 2020 at 16:08
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    $\begingroup$ $xy+yz+xz\ge3*\sqrt[3]{x^2y^2z^2}+2$ isn't true. Your justification for the $+2$ is invalid, as shown by trying $(x,y,z)=(1,2,3)$ $\endgroup$ Aug 11, 2020 at 16:23
  • $\begingroup$ 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? $\endgroup$
    – user814992
    Aug 11, 2020 at 16:39

1 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?

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    $\begingroup$ 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 $\endgroup$
    – user814992
    Aug 11, 2020 at 16:41
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    $\begingroup$ Could you please explain it to me and also thanks a lot for the response $\endgroup$
    – user814992
    Aug 11, 2020 at 16:42
  • $\begingroup$ He just opened the bracket and simplified it $\endgroup$ Aug 11, 2020 at 16:42
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    $\begingroup$ @Michael Christofer I think we can not end your way because AM-GM does not save the case of an equality occurring. $\endgroup$ Aug 11, 2020 at 16:47
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    $\begingroup$ @Michael Christofer You are welcome! Good luck! $\endgroup$ Aug 11, 2020 at 16:49

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