Let $n\in \mathbb{N}$, and $p \in \mathbb{Z}[x]$ be a non-zero polynomial of degree $\le n$. Then, note that:
$$\int_0^1\frac{p(x)}{1+x^2}dx=\frac{A}{n!} + \frac{B\ln(2)}{2}+\frac{C\pi}{4}$$
for some integers $A,B,C \in \mathbb{Z}$. Suppose that $p \ge 0 $ on $[0,1]$. Then:
$$\frac{A}{n!} + \frac{B\ln(2)}{2}+\frac{C\pi}{4} \ge 0$$
It is not difficult to find a necessary and sufficient condition on the coefficients of $p$ for $A,C \neq 0, B=0$. Then, we could rearrange to find a rational inequality for $\pi$, which could be made arbitrarily close if $n$ was allowed to vary. This raises a natural question -- ignoring the constraints on $A,B,C$ for now, which polynomial gives the best bound for fixed $n$?
Question: For $p \in \mathbb{Z}[x]$ with $p \ge 0$ on $[0,1]$ and $\deg p \le n$, is there a polynomial which minimises $\int_0^1\frac{p(x)}{1+x^2}dx$? If so, is it possible to describe a finite set in which this polynomial must lie (in terms of bounds on coefficients)?
I have only proved the trivial lemma:
$$\int_0^1 p(x) dx > 0 \implies \int_0^1 p(x)dx \ge \frac{1}{n!}$$
Intuitively, it is clear that the coefficients of $p$ cannot be too large, but I have found it difficult to make progress on this question because it is hard to consider both the minimisation w.r.t. $\frac{1}{1+x^2}$ and $p \ge 0$ on $[0,1]$ at the same time. I am also not sure if the result holds in general; if $\frac{1}{1+x^2}$ was replaced with an arbitrary weight function $w:[0,1] \to \mathbb{R}$, $w > 0$.
Let me know if a different choice of tags would be more appropriate.