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.



Cyclic inequality for 2 dimensional simplex elements
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.


