<s>[INST] 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 $$f(x)=1 + x + \dots + x^{2k}=\frac{x^{2k+1} - 1}{x-1}.$$
 [/INST] $$p(x) = \frac{2}{ 2k+ 1} ( k + (k+1) x + k x^2 + (k+1)x^3 + \ldots + (k+1) x^{2k-1} + k x^{2k} ) $$
By considering signs (IE Replace $x$ with $-x$), we WTS that for $ x \geq 0$,
$$ k( 1 + x^2 + \ldots + x^{2k}) \geq (k+1)( x + x^3 + \ldots + x^{2k-1}) .$$
Since the number of terms on each side are equal to $k(k+1)$, and the sum of the degrees are the same, it suggests that we can proceed via smoothing out these terms.
This can by done, say by summing up the following inequalities (overall relevant values of $i$):

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*$ \frac{k}{2} x^{2i} + \frac{k}{2} x^{2i+2} \geq k x^{2i+1} $

*

*This gives us $  \frac{k}{2}1 + kx^2 + \ldots + kx^{2k-2} + \frac{k}{2} x^{2k} \geq k( x + x^3 + \ldots + x^{2k-1}) .$



*$ \frac{1}{2} x^0 + \frac{1}{2} x^{2k} \geq \frac{1}{2}x^{2i-1 } +  \frac{1}{2}x^ { 2k-2i+1} $

*

*This gives us $ \frac{k}{2} 1 + \frac{k}{2} x^{2k} \geq ( x + x^3 + \ldots + x^{2k-1})$.



Notes:

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*This can be taught of as essentially Muirhead, since it is clear that $ [ k, 0, k, 0, \ldots, 0, k ] $ majorities $ [ 0, k+1, 0, k+2, \ldots, k+1, 0 ] $.

*Of course, there are other ways that one can do the AM-GM / Jensens.

*Equality holds iff $ x = 1$ (, or $ k = 0$).

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