The missing link: an inequality 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
$$V:=PQ^2P''+(PQ')^2-P^2QQ''-(P'Q)^2.$$
 A: Not an answer, just a plot of $F_n(x)$ for $n=2,\ldots,10$:

         


         


A: This is not an answer, just a reformulation into something which looks perhaps tractable (along the lines of the comment by G. Paseman).  Define, for $m\geq 1$ and $x\in (0,1)$, $$g_m(x):=\frac{d^2}{dx^2}[\log(1+x^m)]=\frac{x^{m-2}(m(m-1)-mx^m)}{(1+x^m)^2}$$ and $$h_m(x):=\frac{d^2}{dx^2}[\log(1-x^m)]=-\frac{x^{m-2}(m(m-1)+mx^m)}{(1-x^m)^2}.$$  Then, for $n\geq 2$, $$F_n''(x)=g_{4n-1}(x)+g_{2n}(x)+h_{2n+1}(x)-g_{2n+1}(x)-h_{2n+2}(x).$$  
Two observations


*

*The contribution of the first, second, and last terms is non-negative for all $n\geq 2$ and $x\in(0,1)$, while that of the remaining terms is non-positive.

*It would therefore suffice to show the following: for $n$ and $x$ as above,
$$g_{2n+1}(x)-h_{2n+1}(x)\leq_? g_{4n-1}(x)+g_{2n}(x)-h_{2n+2}(x).$$


Added:  Since I apparently cannot add a comment, I would just note that while it is certainly (always) possible I overlook something, I have just double checked the algebra with maple, and believe that the expressions are correct (there doesn't seem to be a missing factor of m).  Another note (in addition to the symmetry observed in the comment by Paseman) is the (trivial) identity $h_{2m}=g_m+h_m$, leading to another form for the desired bound:
$$g_{2n+1}(x)+g_{n+1}(x)-h_{2n+1}(x)\leq_? g_{4n-1}(x)+g_{2n}(x)-h_{n+1}(x).$$
A: This is only a comment to @DimaPasechnik, but I cannot put the picture in a comment. The surface to the right of the $x=y$ is a plot of Dima's function (barring mistakes); clearly not convex.

A: I conjecture a stronger statement: that the function $F(x,y)$ of two variables $x,y$ obtained from $F_n(x)$ by substituting $y$ for $x^{2n}$ is convex for $0<y<x<0$. 
EDIT2: this conjecture is wrong as stated. See e.g. the plot in the answer by Yaakov Baruch, or rotate the plot in Sage...
While this reduces to checking positive definiteness of 
a matrix of bivariate polynomials of degree about 20, obtained from the Hessian of $F$ by clearing common (positive) denominator, this should be possible to make to work by standard real algebraic geometry tools.
Here is an (ugly) plot of $F$:
 obtained by exporting the plot from jmol in Sage(math) 
sage: var('x y')
sage: F(x,y)=log((x+y^2)*(1+y)*(1-x*y)/(x*(1+x*y)*(1-x^2*y)))
sage: plot3d(F,[0,1],[0,1])

EDIT: comments say that as $y\to 0+$, the Hessian becomes indefinite. That is, for my original claim to hold, one need to assume $y>\epsilon>0$, with $\epsilon$ possibly depending upon $x$.  
A: I sketch a method that should show $F_n$ is convex for $n$ sufficiently large.
A Taylor expansion gives that $F_n(x) = x^{2n} (1-x)^2 + O(x^{4n-1})$, and so $$F_n''(x) = 2x^{2n-2}[2n^2(1-x)^2 + x^2 - n(1-x)(1+3x)] + O(x^{4n-3}).$$
This should show that $F_n''(x) > 0$ for $n$ large, provided $x < 1-\frac{C}{n}$ for some fixed $C>0$.
On the other hand, according to Mathematica, $F_n''(1) = 4n^2 - \frac{16}{3} n + \frac{5}{4}$, which has its largest root at $\frac{8 + \sqrt{19}}{12} \approx 1.02991$.  (By the way, there doesn't seem to be any reason that $n$ should be restricted to be an integer, and this is in line with some plots I made indicating $F_n(x)$ is convex for $n > 1.03$.)
Moreover, Mathematica gives $$\lim_{n\rightarrow \infty} n^{-2}F_n''(1-\frac{y}{n}) = \frac{16 e^{4y}}{(1+e^{4y})^2} > 0.$$
Putting these two pieces together should be able to show $F_n(x)$ is convex on $(0,1)$ for $n$ sufficiently large.  Unfortunately, it might be very painful to work out all the error terms explicitly.
