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 $$f(x)=1 + x + \dots + x^{2k}=\frac{x^{2k+1} - 1}{x-1}.$$