My question is very simple (I think so), but I have a trouble to get the precise understanding of the following question.
Let's take into consideration the following function: $$f(n,a,s_1,s_2,...s_n)=\frac{1}{n(1+s_1)}+\frac{1}{n(1+s_2)^2}+...+\frac{1}{n(1+s_n)^n}-\left(\frac{a}{n(1+s_2)^2}+\frac{2a}{n(1+s_3)^3}+...\frac{(n-1)a}{n(1+s_n)^n}\right)-\frac{1}{(1+s_n)^n},$$ where $n$ is integer, $a\in(0,1), s_i\in(0,1), i=1, 2,...,n$ and $s_1<s_2<...<s_n$. The function $f$ attains positive, negative as well as zero values.
If we replace all $s_i$'s $(i=1,...,n-1)$ by $s_n$ is it correct to write the following relationship (note: $s_1<...<s_n$) $$\frac{1}{n(1+s_1)}+\frac{1}{n(1+s_2)^2}+...+\frac{1}{n(1+s_n)^n}-\left(\frac{a}{n(1+s_2)^2}+\frac{2a}{n(1+s_3)^3}+\frac{(n-1)a}{n(1+s_n)^n}\right)-\frac{1}{(1+s_n)^n}>\frac{1}{n(1+s_n)}+\frac{1}{n(1+s_n)^2}+...+\frac{1}{n(1+s_n)^n}-\left(\frac{a}{n(1+s_n)^2}+\frac{2a}{n(1+s_n)^3}+...\frac{(n-1)a}{n(1+s_n)^n}\right)-\frac{1}{(1+s_n)^n}.$$
It is easy to check that the inequality you asking is equivalent to
$$g(n,a,s_1,s_2,\dots,s_n)=\sum_{i=1}^n (1-(i-1)a)\left(\frac{1}{(1+s_i)^i}-\frac{1}{(1+s_n)^i}\right)>0.$$
This may be wrong for $a$ close to 1. For instance, MathCad calculates $g(4,0.99,0.05,0.1,0.15,0.9)\simeq –0.07$.