Consider the following sequence defined for a fixed number $ n \geq 2, n \in \mathbb{N}$: Let $c_{1}, c_{2}, ..., c_{n} \in \mathbb{R}_{>0}$ be positive real numbers. Let $T_{i}$ denote a tuple containing product of i items, that is, $$ \begin{aligned} T_{1} &= \{c_{1}, c_{2}, \ldots, c_{n} \} \\ T_{2} &= \{c_{1}c_{2}, c_{1}c_{3}, \ldots, c_{2}c_{3}, \ldots,\} \\ &\vdots \\ T_{n-1} &= \{c_{1}c_{2}c_{3} \cdots c_{n-1}, c_{1}c_{2}\cdots c_{n-2}c_{n}, \ldots, c_{2} \cdots c_{n-2}c_{n}\}\\ T_{n} &= \{c_{1}c_{2}c_{3} \cdots c_{n-1}c_{n} \} \end{aligned} $$
The sequence is defined as follows for $i = 1, 2, ..., n$:
$$s_{i} = \frac{(n-(i-1)) \sum_{t_{p} \in T_{n-(i-1)}} t_{p}}{i \sum_{t_{q} \in T_{n-i}} t_{q}} \, .$$
Prove that the sequence $s_{i}$ is monotonic and non-decreasing for a given $n$, that is, $s_{1} \leq s_{2} \leq ... \leq s_{n}$.
For example, when $n=2$, $s_{1} = \frac{2 c_{1} c_{2}}{c_{1} + c_{2}}$, $s_{2} = \frac{c_{1} + c_{2}}{2} $. Clearly, $s_{1} \leq s_{2}$.
I tried to prove by showing $s_{i} \leq s_{i+1}$. That means,
$$\frac{(n-(i-1)) \sum_{t_{p} \in T_{n-(i-1)}} t_{p}}{i \sum_{t_{q} \in T_{n-i}} t_{q}} \leq \frac{(n-i) \sum_{t_{q} \in T_{n-i}} t_{q}}{(i+1) \sum_{t_{r} \in T_{n-(i+1)}} t_{r}}$$ or:
$$(n-i)i (\sum_{t_{q} \in T_{n-i}} t_{q})^{2} \geq (n-(i-1)) (i+1) (\sum_{t_{p} \in T_{n-(i-1)}} t_{p}) (\sum_{t_{r} \in T_{n-(i+1)}} t_{r}) \, .$$
If I substitute $n$ as 3 or 4 and $i$ as 1 or 2 , I am able to simplify and complete the square and prove its true, but how do I show it generally?
You want to show that $$ (n-i)i (\sum_{t_{q} \in T_{n-i}} t_{q})^{2} \geq (n-(i-1)) (i+1) (\sum_{t_{p} \in T_{n-(i-1)}} t_{p}) (\sum_{t_{r} \in T_{n-(i+1)}} t_{r}) \, . $$ This can be written as $$ (n-i)i \sigma_{n-i}^2 \ge (n-i+1)(i+1) \sigma_{n-i+1} \sigma_{n-i-1} $$ where $\sigma_k$ is the $k$th elementary symmetric function in $c_1, \ldots, c_n$. Now replace $i$ by $n-i$ to simplify the notation: $$ i(n-i) \sigma_i^2 \ge (i+1)(n-i+1) \sigma_{i+1} \sigma_{i-1} \, . $$ This is equivalent to $$ \frac{\sigma_i^2}{{\binom ni}^2} \ge \frac{\sigma_{i+i}}{\binom n{i+1}} \cdot \frac{\sigma_{i-i}}{\binom n{i-1}} $$ and that are exactly Newton's inequalities.