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This question accross to this question from SE which there some answers but they r n't

enough to me hop to see MO what can they say about it .

let $m,n$ be integers, show that if $ n>m\geq 0 $ :

$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$

where real $x,y,z > 0 $ and $xy + yz + zx = 1$

Note : The question is Already montioned here in journal k.s competition prolem 111.

Edit :The choice of $x, y ,z$ gaven by peterMuller, does not fulfill the auxiliary condition $xy+yz+zx=1$ as it is required in the problem, so it cannot serve as a counter example. In any case, even with him choice of $x,y,z,$ he still need to show it is less than $\sqrt{3}/2$, since this is the claim in the problem

Thank you for your help .

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    $\begingroup$ Where did you find this inequality? $\endgroup$
    – GH from MO
    Commented Sep 15, 2015 at 21:10
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    $\begingroup$ Might look a bit less messy if you let $X=\sqrt3x$, $Y=\sqrt3y$, and $Z=\sqrt3z$. That gets rid of the $(1/\sqrt3)^{n-m}$ on the right, at the cost of changing the boundary condition to $XY+YZ+ZX=3$. $\endgroup$
    – Joe Silverman
    Commented Sep 15, 2015 at 22:04
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    $\begingroup$ Feels like a math competition inequality... $\endgroup$
    – Anthony Quas
    Commented Sep 16, 2015 at 0:01
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    $\begingroup$ Thank you for waiting until after the date for submitting solutions before posting here. $\endgroup$
    – Gerald Edgar
    Commented Sep 16, 2015 at 15:34
  • $\begingroup$ @GeraldEdgar, but why this downvote to my uestion such that i hput it source ? $\endgroup$
    – zeraoulia rafik
    Commented Sep 16, 2015 at 22:27

1 Answer 1

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This inequality doesn't hold in general. It is false for instance for $m=7$, $n=8$: Set $X=3/4$, $Y=1$, and $Z=9/7$. Then $XY+YZ+ZX=3$, however $\frac{X^8}{X^7+Y^7}+\frac{Y^8}{Y^7+Z^7}+\frac{Z^8}{Z^7+X^7}<\frac{3}{2}$.

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    $\begingroup$ Aha, so the math competition problem there isn't correct. By the way, you should have told the source of this question, and given a reason to post it at MO. $\endgroup$
    – Peter Mueller
    Commented Sep 16, 2015 at 13:10
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    $\begingroup$ Peter is right. We generally don't do competition problems, and there are good reasons we don't. Thus if you submit a competition problem, you should always disclose the source of the problem and explain why an exception should be made, as a courtesy. You were even asked earlier by GH from MO but decided not to respond; why? $\endgroup$
    – Todd Trimble
    Commented Sep 16, 2015 at 17:30
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    $\begingroup$ @ToddTrimble,sorry i was offline for giving a response to GHfrom MO about source of this question and i posted it in MO after the date of submitting it to journal k.s. $\endgroup$
    – zeraoulia rafik
    Commented Sep 16, 2015 at 20:18
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    $\begingroup$ If you're wondering why I called it a "competition problem": it's a "competition" to see if one can solve the problem by one's self and get one's name in the journal; the "winners" are those that do, and first prize is to have one's solution be the one published as the official solution. We don't discuss that kind of thing here except in rare circumstances (usually because it ties in with a bona fide research question). $\endgroup$
    – Todd Trimble
    Commented Sep 16, 2015 at 21:37
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    $\begingroup$ @zeraouliarafik: Are you really unable to check if my examples are counter examples or not? At math.stackexchange.com/questions/1306593/… I've added a second dis-proof which does not require to compute with large integers. $\endgroup$
    – Peter Mueller
    Commented Dec 14, 2015 at 10:40

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