Polynomial inequality: show $p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n\ge 0$ Consider the following polynomial in $x$:
$$p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n.$$
I want to show that $p(x)\geq 0$ for real $x$. It is trivial to show that $p(-1) = 0$, and by graphing $p(x)$ for some $k$ we can see that $x=-1$ is in fact the global minimum of this polynomial, presumably for any $k\in \mathbb{N}^+$.
How can I show that this is in fact true?
Note that $$p(x)=f(x) - \frac{1}{2k+1}f(-x),$$ with $$f(x)=1 + x + \dots + x^{2k}=\frac{x^{2k+1} - 1}{x-1}.$$

Changing your notation slightly, we have
$$
p_{2k}(x) = \frac{2}{2k+1}\left(b_0 + b_1 x + b_1 x^2 + \cdots + b_{2k}x^{2k}+ b_{2k+1} x^{2x+1}\right),
$$where $b_n = k$ if $n$ is even and $b_n=k+1$ if $n$ is odd. We will prove the claim by induction on $k$. The base case $n=0$ gives $p_0(x) =0$ and we also have $p_2(x) = 2/3 (x+1)^2$, which is obviously non-negative.
Suppose $p_{2k}(x) \ge 0$, with a minimum at $x=-1$. We have
$$
p_{2k+2}(x)- p_{2k}(x) = (x+1)^2\cdot \frac{2}{4k^2-1}\sum_{n=0}^{2k-2} (-1)^n\frac{(n+1)(n+2)}{2} x^n
$$Cancelling with the factor of $(x+1)^2$, the sum reduces to
$$
S_{2k}(x)=\frac{2 k^2 (x + 1)^2 x^{2k-1} - k (x^2 - 1) x^{2k-1} -x^{2k} + 1}{ x + 1};$$ note that $S_{2k}(0)=1$ and $\lim_{x\to -1} S_{2k}(x) $ exists and is positive by LHR.
We claim the numerator has a unique root at $x=-1$. Indeed,
$$
2 k^2 (x + 1)^2 x^{2k-1} - k (x^2 - 1) x^{2k-1} -x^{2k} + 1$$
$$ = (2k^2-k)x^{2k+1}+(4k^2-1)x^{2k}+(2k^2+k)x^{2k-1}+1,
$$
and its derivative is
$$
x^{2k-2}\bigg((2k^2-k)(2k+1)x^2 + (4k^2-1)(2k)x + (2k^2+k)(2k-1)\bigg)
$$By the quadratic formula, this has a unique double root at $x=-1$. This means the numerator of $S_{2k}$ changes sign at $x=-1$; in particular, the numerator is negative for $x<-1$ and positive for $x>-1$, implying $S_{2k}(x)>0$. In summary, $p_{2k}(x)$ is clearly zero at $x=-1$ and is increasing in $k$ for any fixed $x$, so the inequality holds.
