Show that $\sum\limits_{k=1}^{n-1} (n-k) x^k$ is non-decreasing for $x \in ]-1,1[$. Question
I would like to show for arbitrary $n \in \mathbb{N}$, that the polynomial:
$$
p(x) := \sum_{k=1}^{n-1} (n-k) x^k
$$
is non-decreasing.
Start Of Solution
We can write its derivative by:
$$
p'(x) = \frac{(-(x + 1) (x^n-1) + n (x - 1) (x^n + 1))}{(x - 1)^3},
$$
as $(-1 + x)^3 < 0$ for $x \in ]-1,1[$ it suffices to show for arbitrary $n$ that:
$$
n(x-1) (x^n + 1) \leq (x+1)(x^n - 1),
$$
for $n = 1$ this is trivial (as both sides become equal). Thus we assume the inequality to hold for $n$ and show it for $n+1$, here I use $x^{n+1} + 1 = x(x^n+1) + (1-x)$ to use the induction hypothesis but I don't achieve the inequality through this method.
No positive roots
By Descartes' Sign rule we see that $p'(x) = \sum\limits_{k=1}^{n-1}(n-k)k x^{k-1}$ has no positive roots, thus it suffices to show 
 A: I found it I think.
Let's replace $n$ by $n+1$ to obtain:
$$
p(x) = \sum_{k=1}^n(n+1-k)x^k,
$$
now we let the sum go from $0$ to $n-1$:
$$
p(x) = \sum_{k=0}^{n-1} (n-k)x^{k+1},
$$
taking the derivative yields:
$$
p'(x) = \sum_{k=0}^{n-1}(n-k)(k+1)x^k
$$
it now suffices to show that $p'(x) \geq 0$ on $]-1,1[$.
One can check (for example with mathematica) that we have:
$$
p'(x) = \frac{n(x^{n+1} + 1) - 2 (x^n + \dots + x))}{(x-1)^2},
$$
thus we just need to show that 
$$\frac{(x^{n+1} + 1)}{2} \leq \frac{x^n+\dots+x}{n}.$$
For $x \in ]-1,0]$ we have  $\frac{(x^{n+1} + 1)}{2} \geq 0$ while $ \frac{x^n+\dots+x}{n} = x(1-x^n)/(1-x) \leq 0$, so we may assume $x \in [0,1[$. We do this part by induction skipping the case $n = 1$. We have:
$$
\frac{x^{n+1} + \dots + x}{n+1} = \frac{nx \frac{x^n + \dots + x}{n} + x}{n+1} \leq \frac{nx \frac{x^{n+1} + 1}{2}+x}{n+1} \overset{?}{\leq} \frac{x^{n+2} + 1}{2}.
$$
It remains to show that the last inequality holds. Subtracting the left hand side from the right hand side we obtain:
$$
\frac{(x^{n+2} - ((n+2)x - n-1)}{2(m+1)},
$$
which is positive as $x^{n+2}$ is convex and the tangent line at $1$ is given by $y = ((n+2)x - n -1)$.
This completes the proof.
