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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][1] 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 $\tau$.

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. [1]: Technical lemma on root systems, reduced to linear algebra

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 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 $\tau$.

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.

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][1] 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 $\tau$.

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. [1]: Technical lemma on root systems, reduced to linear algebra

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][1] 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 $\tau$.

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. [1]: Technical lemma on root systems, reduced to linear algebra

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 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 $\tau$.

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][1] 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 $\tau$.

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. [1]: Technical lemma on root systems, reduced to linear algebra