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 $$ \frac{b_1 + b_2 + \cdots + b_i + b_{\tau(1)} + \cdots + b_{\tau(i)}}{2i} \geq \frac{b_1 + b_2 + \cdots + b_n}{n} \quad \dots \text{Eq. }(i)$$ 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 $\tau$.