I've been working on a project and proved a few relevant results, but got stuck on one tricky problem:
Conjecture. If $2\leq n\in\mathbb{N}$ and $0<x<1$ is a real number, then
$$F_n(x)=\log\left(\frac{(1+x^{4n-1})(1+x^{2n})(1-x^{2n+1})}{(1+x^{2n+1})(1-x^{2n+2})}\right)$$
is a convex function of $x$.I have a heuristic argument. Can you help with a rigorous proof or valuable tools?
Further motivation. If you succeed with this, then I'll be honored to have you as a co-author in this work. The problem itself can be found in Section 4.
Note. Write $F=\log\frac{P}Q$, then $F''>0$ amounts to the positivity of the polynomial
$$V:=PQ^2P''+(PQ')^2-P^2QQ''-(P'Q)^2.$$
Best Answer
Fix $n$, and let $V(x)$ be the polynomial from the question, for which we want to show that it is positive on the open interval $(0,1)$. Set \begin{align*} a(n) &= 1024n^2 - 4096/3n + 320\\ b(n) &= 12288n^3 - 14336n^2 + 4864/3n + 256\\ W(x) &= x^4\left(V(x)-(n+1)^2(2n+1)^2x^{24n-1}(1-x)^4(a(n)x + b(n)(1-x))\right). \end{align*}
For all $n\ge12$ we claim that $(1-x)^6$ divides $W(x)$, and that all the coefficients of $W(x)/(1-x)^6$ are nonnegative. In particular, $W(x)$ is positive for $x>0$. As $a(n)$ and $b(n)$ is positive for all $n\ge1$, we see that $V(x)$ is positive for all $0<x<1$, provided that $n\ge12$. The cases $2\le n\le 11$ are easily checked directly, for instance by noting that the coefficients of $(1+x)^{\deg V}V(1/(1+x))$ are positive.
I'm not sure about the easiest kind to prove the assertion about the coefficients of $W(x)/(1-x)^6$. We compute them via \begin{equation} \frac{W(x)}{(1-x)^6}=W(x)\sum_{k\ge0}\binom{k+5}{5}x^k. \end{equation}
All what follows is assisted/confirmed by the Sage code below.
The exponents of $W(x)$ have the form $2ni+j$ for $0\le i\le 12$ and $0\le j\le 10$. Let $a_{i,j}$ be the corresponding coefficient. It is a polynomial in $n$.
We have \begin{equation} \frac{W(x)}{(1-x)^6} = \sum_{i,j}a_{i,j}(n)x^{2ni+j}\sum_k\binom{k+5}{5}x^k. \end{equation}
With $r\ge0$ and $0\le s\le 2n-1$, the coefficient of $x^{2nr+s}$ in $W(x)\frac{1}{(1-x)^6}$ is the sum of all $a_{i,j}(n)\cdot\binom{nr+s-ni-j+y}{5}$ for all pairs $(i,j)$ where $i<r$ and $0\le j\le 10$, or $i=r$ and $j\le s$.
We distinguish the cases $0\le s\le 9$ from the cases $s\ge10$. In these latter cases, we write $s=10+y$. Recall that $s\le 2n-1$, so $10+y=2n-1-o$ for a nonnegative $o$. Upon replacing $n$ with $(11+y+1)/2$, we get polynomials in $n$ and $o$.
In the former cases, the coefficients are polynomials in $n$, which have positive coefficients upon replacing $n$ with $n+12$. So these coefficients are positive for all $n\ge12$.
In the latter cases it turns out that in all but one case the coefficients are positive. So for nonnegative $y$ and $o$, the values are nonnegative.
For the single exception we see that if we multiply it with a suitable polynomial, the resulting coefficients are positive.
Remark (answering Jason's question): It is a known fact that a polynomial $V$ is positive on $(0,1)$ if and only if it is a nonnegative linear combination of polynomials $x^i(1-x)^j$. The problem is that it may involve terms where $i+j$ is bigger than $\deg V$. (This doesn't happen here, though.) I used an LP solver to play a little with $V$ and $\frac{V}{(1-x^2)^4x^{2n-2}}$. From that a pattern showed up which lead to a solution.
By the way, $V$ actually seems to be a nonnegative linear combination of $x^i(1-x)^{\deg V-i}$. This is equivalent to say that all the coefficients of $(1+x)^{\deg V}V(1/(1+x))$ are nonnegative. It is not hard to compute explicit expressions for these coefficients, but I don't see an easy arguments why they can't be negative.