Let $p=(p_{1},p_{2},p_{3})\in\Delta$, with $\Delta:=\lbrace p\in(0,1)^{3}\ |\ p_{1}+p_{2}+p_{3}=1 \rbrace$. I aim to prove (not knowing whether it is true though) that \begin{equation} p_{1}^{p_{3}-p_{2}}p_{2}^{p_{1}-p_{3}}p_{3}^{p_{2}-p_{1}}\le1. \end{equation} Indeed, if at least two of the three numbers are equal, then the inequality holds (with equality) (thus we may assume wlog $p_{1}<p_{2}<p_{3}$). I've tried a plenty of examples and couldn't find any for which it is wrong. Yet, I fail to prove the validity of this inequality so far and I am therefore thankful for any help.
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$\begingroup$ Then why $(p_{1}-p_{3})\ln p_{2}\le (p_{2}-p_{3})\ln p_{1} +(p_{1}-p_{2})\ln p_{3}$ ? $\endgroup$– TobsnCommented May 13, 2020 at 12:22
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$\begingroup$ Still not sure if I get your point. If we consider $p_{1}<p_{2}<p_{3}$ then in $(p_{3}-p_{2})\ln p_{1}+(p_{1}-p_{3})\ln p_{2} +(p_{2}-p_{1})\ln p_{3}$ clearly the first and third term are negative, I agree, the second term however is positive, so why is the whole term negative? $\endgroup$– TobsnCommented May 13, 2020 at 12:47
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I don't think your 'wlog' is correct. I think you can only assume 3 distinct, and not an order. If you agree with this, then rewriting as \begin{equation} p_{1}^{p_{3}}p_{2}^{p_{1}}p_{3}^{p_{2}}/ p_{1}^{p_{2}}p_{2}^{p_{3}}p_{3}^{p_{1}} \end{equation} if you call the numerator $f(p_1,p_2,p_3)$ the the denominator is $f(p_3,p_2,p_1)$ and it can't be bigger in general.
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$\begingroup$ Gosh, thats true. Unfortunately :D. So then, probably the statement works only on the ordered simplex. But even if, I'm not sure whether this would still be helpful for my purposes. Anyway, thank you! $\endgroup$– TobsnCommented May 13, 2020 at 13:10
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If $p_1=3/100$, $p_2=77/100$, and $p_3=20/100$, then the left-hand side of your inequality is $2.3447\ldots>1$. So, your inequality is false in general.
answered May 13, 2020 at 13:09