Proving an inequality involving discrete variables I'm trying to show that the following inequality holds
$$
\frac{1-x^{n}}{1-x^{n+1}}\geq\frac{\sum_{i=0}^{n-2}x^{i}(1-x_{1}^{n-(i+1)})}{\sum_{i=0}^{n-1}x^{i}(1-x_{1}^{n-i})},
$$
where $n$ is a positive integer, $x$ and $x_{1}$ are real numbers
satisfying $x_{1}<1$ and $x>x_{1}$. 
This is what I have tried: One can show that the left-hand-side
is increasing in $n$, hence to show the above it suffices to show that
$$
\frac{\sum_{i=0}^{n-2}x^{i}}{\sum_{i=0}^{n-1}x^{i}}\geq\frac{\sum_{i=0}^{n-2}x^{i}(1-x_{1}^{n-(i+1)})}{\sum_{i=0}^{n-1}x^{i}(1-x_{1}^{n-i})}.
$$
I have tried bounding the terms $(1-x_{1}^{n-(i+1)})$ and $(1-x_{1}^{n-i})$
in the right-hand-side and bring them out of the sum but that doesn't
lead to the desired result.
 A: Evaluating the four geometrical sums in 
$$f_n(x,x_1) = (1-x^n)\sum_{i=0}^{n-1}x^{i}(1-x_{1}^{n-i}) - (1-x^{n+1})\sum_{i=0}^{n-2}x^{i}(1-x_{1}^{n-(i+1)})$$
and simplifying we obtain
$$f_n(x,x_1) = (x_1x)^n + \frac{g_n(x) - g_n(x_1)}{x-x_1}$$
where $g_n(x) = x^n(1-x)$. From this expression we see that the function $f_n$ is symmetric $f_n(x,x_1)=f_n(x_1,x)$ so the condition $x>x_1$ should not be (and is not) needed.
We will show that $(1-x_1)(1-x)f_n\geq 0$ for all $x,x_1\geq 0$ where the prefactors $(1-x)(1-x_1)$ have been introduced to avoid splitting into different cases for $x>1$ and $x<1$ etc. To do this we use induction on $n$. Assume that for $n\geq 2$ we have  $(1-x_1)(1-x)f_n\geq 0$ then
$$(1-x_1)(1-x)f_{n+1} = (1-x)^2x^n(1-x_1)[1-x_1^{n+1}] + x_1(1-x_1)(1-x)f_n \geq 0$$
For the base case $n=2$ we can factorize $(1-x_1)(1-x)f_2$ as
$$(1-x)f_2 = (1-x)^2(1-x_1)^2\left[x + (1+x)(1+x_1)\right] \geq 0$$
and it follows that the inequality
$$\frac{(1-x)(1-x_1)f_n(x,x_1)}{(1-x)(1-x_1)(1-x^{n+1})\sum_{i=0}^{n-2}x^i(1-x^{n-i-1})} \geq 0 \implies \frac{1-x^n}{1-x^{n+1}} \geq \frac{\sum_{i=0}^{n-1}x^{i}(1-x_{1}^{n-i})}{\sum_{i=0}^{n-2}x^{i}(1-x_{1}^{n-(i+1)})}$$
holds for all $x,x_1\geq 0$. The range of allowed values can in fact be sharpened to  $x \geq -\frac{1+x_1}{2+x_1}$, $x_1\geq 0$ and by symmetry the inequality also holds for all $x_1\geq -\frac{1+x_1}{2+x_1}$ and $x\geq 0$.
