Title
Generalizing $a^ab^b>a^bb^a$, $a^ab^bc^c>a^bb^cc^a$.
Background
We consider two distinct numbers $a,b\in\mathbb{R}^+$, no matter $a<b$ or $b<a$, we have: $$\frac{a^ab^b}{a^bb^a}=\Big(\frac{a}{b}\Big)^{a-b}>1$$ and hence $a^ab^b>a^bb^a$.
I am wondering about possible generalization(s). First of all, for the case of three distinct numbers $a,b,c\in\mathbb{R}^+$, $a^ab^bc^c>a^bb^cc^a$ and $a^ab^bc^c>a^cb^ac^b$, or simply $a^ab^bc^c>\max\{a^bb^cc^a,a^cb^ac^b\}$. To prove this, assume $a>b>c$, and notice that $$\frac{a^ab^bc^c}{a^bb^cc^a}>\frac{a^bb^ac^c}{a^bb^cc^a}=\frac{b^ac^c}{b^cc^a}=\Big(\frac{b}{c}\Big)^{a-c}>1$$ and $$\frac{a^ab^bc^c}{a^cb^ac^b}>\frac{a^bb^ac^c}{a^cb^ac^b}=\frac{a^bc^c}{a^cc^b}=\Big(\frac{a}{c}\Big)^{b-c}>1.$$ If mutually distinct $a,b,c\in\mathbb{R}^+$ in general, denoted by $a_0,a_1,a_2$ respectively, we could let bijective mapping $\sigma:\{0,1,2\}\to\{0,1,2\}$ be a permutation of 0,1,2 such that $a_{\sigma(0)}>a_{\sigma(1)}>a_{\sigma(2)}$ so that now $$a_0^{a_0}a_1^{a_1}a_2^{a_2}=a_{\sigma(0)}^{a_{\sigma(0)}}a_{\sigma(1)}^{a_{\sigma(1)}}a_{\sigma(2)}^{a_{\sigma(2)}}>a_0^{a_{(0\space+_3\space k)}}a_1^{a_{(1\space+_3\space k)}}a_2^{a_{(2\space+_3\space k)}}$$ where $k=1$ or $2$ and $+_3$ denotes the addition in $\mathbb{Z}/3\mathbb{Z}$. Notice that this is possible because at the rightmost expression the exponent of $a_i$ is $a_j$ for some $j\neq i$ and notice that for $n=3$, if a bijective mapping $\sigma:\{0,1,2\}\to\{0,1,2\}$ has the property $\sigma(i)\neq i$ for $i=0,1,2$, then (by listing out all permutations of $(1,2,3)$) we can lead to the conclusion that $\sigma(i)=(i\space+_3\space k)\space\forall i=0,1,2$ for $k=1$ or $2$.
My conjecture
For $n\in\mathbb{N}$, for $a_0,\cdots,a_{n-1}$, must we have $$\prod_{i=0}^{n-1}a_i^{a_i}>\prod_{i=0}^{n-1}a_i^{a_{\sigma(i)}}$$ where $\sigma:\{0,\cdots,n-1\}\to\{0,\cdots,n-1\}$ is any bijective map such that $\exists i: \sigma(i)\neq i$. Also, if we relax the condition that $\exists i: \sigma(i)\neq i$, maybe we can replace the $>$ by $\ge$ and the inequality still holds? Notice that the conjecture says that product of at least 2 power of mutually distinct positive real numbers to itself will be strictly reduced when the exponents are re-arranged, which is quite interesting to my point of view.
More about it
I have tried to generalize the case for $n>3$ by considering a sequence of swapping so that $(0,\cdots,n-1)$ is changed to $\big(\sigma(0),\cdots,\sigma(n-1)\big)$, and hoping that we can use $a^ab^b>a^bb^a$ to prove the conjecture step by step. However, consider the following example:
Example making me frustrated:
For $n=6$, let $\sigma(i)=5,2,3,1,0,4$ for $i=0,1,\cdots,5$ respectively. Denote mutually distinct positive real numbers $a_0,\cdots,a_{5}$ by $a,b,c,d,e,f$ respectively, and we hope that we can prove:
$$a^ab^bc^cd^de^ef^f>a^fb^cc^dd^be^af^e.$$
We may first swap the 4-th and 5-th terms so that $a^ab^bc^cd^de^ef^f>a^ab^bc^cd^de^ff^e$.
We may then swap the 2-nd and 3-rd terms so that $a^ab^bc^cd^de^ef^f>a^ab^bc^dd^ce^ff^e$ now.
We may then swap the 1-st and 3-rd terms so that $a^ab^bc^cd^de^ef^f>\cdots(?)$
: The problem is that I thought I could have $b^bd^d>b^dd^b$ to use but the power of $d$ in the previous step is already changed. So I can't prove the conjecture I proposed.
Conclusion
Is it possible to prove my conjecture easily? Of course, things may not always be that easy, but I am not experienced enough in every subject in Mathematics, so please state where I can find the knowledge required clearly (something like permutation?) if possible. Thank you a lot!
At the end, I will appreciate any kind of help!
One can assume that the $a_i$ and $b_i = \ln(a_i)$ are both sorted in (weakly) increasing order. Then $$ \ln \left(\prod_{i=0}^{n-1}a_i^{a_{\sigma(i)}} \right) = \sum_{i=0}^{n-1} a_{\sigma(i)} b_i \le \sum_{i=0}^{n-1} a_{i} b_i $$ follows from the rearrangement inequality. If the $a_i$ are pairwise distinct then equality holds only if $\sigma$ is the identity permutation.