They say that all mathematics problems eventually reduce to linear algebra or combinatorics. I have reduced mine to proving a solution exists for the following set of inequalities but have no idea how to proceed.


**Question:** For $n \geq 2$, fix once and for all, a permutation $\tau \in S\_n, \tau \neq (1,n)(2,n-1)\cdots$, i.e., $\tau$ isn't the long element. Let
$$ \Delta(\tau) = \{ i \in \{1, 2, \cdots, n-1 \} : \tau(\{ 1, 2, \cdots, i \}) \neq \{ n, n-1, \cdots, n-i+1 \} \}$$


Do there exist real numbers $b\_1 > b\_2 > \cdots > b\_{n-1} > b\_n = 0$ such that the inequalities


$$ \frac{b\_1 + b\_2 + \cdots + b\_i + b\_{\tau(1)} + \cdots + b\_{\tau(i)}}{2i} > \frac{b\_1 + b\_2 + \cdots + b\_n}{n} \quad \dots \text{Eq. }(i)$$
hold simultaneously for every $i \in \Delta(\tau)$?


**Motiation:** My original question is posted [here](https://mathoverflow.net/questions/259765/) and the above question is the case when $G = SL(n)$, after some simplifications and explicit computation with the Cartan matrix.


**Remark:** I have explicitly verified this holds for $n \leq 8$. These calculations suggest that the $b\_i$'s cannot be independent of $\tau$.




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> 
> **MAJOR EDIT:** I made a mistake while posing the question. The $\tau$ in Equation $(i)$ should actually be $\tau^{-1}$. The mistake came
> because $\sum b\_i \varpi\_{\tau(i)}$ was wrongly written as
> $\sum b\_{\tau(i)} \varpi\_i$ instead of $\sum b\_{\tau^{-1}(i)} \varpi\_i$. I will reward the bounty to the
> answer by @Pietro Majer but the question I really want answered is:
> 
> 
> 


**Do there exist real numbers $b\_1 > b\_2 > \cdots > b\_{n-1} > b\_n = 0$ such that the inequalities
$$ \frac{b\_1 + b\_2 + \cdots + b\_i + b\_{\tau^{-1}(1)} + \cdots + b\_{\tau^{-1}(i)}}{2i} > \frac{b\_1 + b\_2 + \cdots + b\_n}{n} \quad \dots \text{New Eq. }(i)$$hold simultaneously for every $i \in \Delta(\tau)$?**



> 
> I can award some more bounty points for proving the new inequality.
> 
> 
> 



Existence of solution to these inequalities
For a given $\tau\in S\_n$ and for any $j\in[n]$ define the numbers
$$a\_j:=\chi\_{\Delta}(j)-\chi\_{\Delta}(j-1)-\chi\_{\Delta }(n-j)+\chi\_{\Delta}(n-j+1),$$
where $\chi\_\Delta:\mathbb{Z}\to\{0,1\}$ denotes the characteristic function of the set $\Delta:=\Delta(\tau)\subset\mathbb{Z}$. Also, with $c:=5n-a\_n$, define
$$b\_j:=a\_j-5j+c.$$
We may note right away that since $|a\_j|\le2$, the $b\_j$ are strictly decreasing, and that $b\_n=0$, by the choice of the constant $c$.


For any $E\subset[n]$, for simplicity of notation we put
 $$ \alpha(E):=\sum\_{j\in E} a\_j,\qquad \beta(E):=\sum\_{j\in E} b\_j$$
(so we may think $ \alpha$ and $\beta$ as discrete signed measures supported in $[n]$).


For $i\in[n]$, summing over $j=1,\dots i$ we have


$$ \alpha([i])=\chi\_\Delta(i)-\chi\_\Delta(n-i).$$


Incidentally, for any $i\in[n]$ we have $i\in\Delta$ if and only if, by definition, $\tau([i])\neq[i]+n-i$ thus also, since $\tau$ is bijective, if and only if $\tau([i]^c)\neq([i]+n-i)^c$, that is $\tau([n-i]+i)\neq [n-i]$ or $\tau^{-1}([n-i])\neq [n-i]+i$, which means $n-i \in \Delta^{-1}:=\Delta(\tau^{-1})$. Hence the last formula also writes
$$ \alpha([i])=\chi\_{\Delta}(i)-\chi\_{\Delta^{-1}}(i).$$
Also note that, since $n\not\in\Delta$
$$ \alpha([n])=0,$$
and $$ \alpha([i]+n-i)=- \alpha([n-i]) =-\chi\_\Delta(n-i)+\chi\_\Delta(i)= \alpha([i]).$$


We proceed showing the inequalities on the arithmetic means. 


**Case I.** Assume $i\in\Delta\setminus\Delta^{-1}$. Then by definition of $\Delta^{-1}$, $\tau^{-1}([i])=[i]+n-i$, so that


$$ {\alpha([i])+ \alpha(\tau^{-1}[i])\over 2i}= {\alpha([i])+ \alpha([i]+n-i)\over2i}={\chi\_{\Delta}(i)-\chi\_{\Delta^{-1}}(i)\over i}={1\over i}>0, $$
and summing the arithmetic means of $-5j+c$ on the same sets we have plainly
$${\beta([i])+ \beta(\tau^{-1}[i])\over 2i}>{\beta([n])\over n}.$$


**Case II.** Assume $i\in\Delta\cap\Delta^{-1}$. Thus $\tau^{-1}([i])\neq[i]+n-i$ and, just because $b\_j$ are strictly decreasing


$${\beta([i])+ \beta(\tau^{-1}[i])\over 2i}>{\beta([i])+ \beta([i]+n-i)\over2i}$$
and since we have $\alpha([i])=\alpha([i]+n-i)=\alpha([n])=0$ because $\chi\_{\Delta}(i)=\chi\_{\Delta^{-1}}(i)=1$, summing as before the arithmetic means of the affine part of $b\_j$, $${\beta([i])+ \beta([i]+n-i)\over2i}={\beta([n])\over n},$$
concluding the proof.


