Time to start typing. As I said, this is going to be long and boring and I have only limited amount of free time nowadays so I'll type in chunks. If you see a flaw somewhere, comment immediately, but if everything looks like a chain of correct stupid computations leading nowhere, just keep patient and wait until it is finished.

Before passing to the proof, I should say that I slightly disagree with Pietro on what is the main issue with this problem. IMHO, it is its non-generic nature: some combinations like that are convex and some are not. Moreover, the really beautiful and structured ones aren't for simple symmetry reasons. So, I do not believe there is any good underlying mechanism for the result to be true. It is just correct by an accident and grows in the mathematical forest like an occasional mushroom that is there for no apparent reason. You can still pick it up if you can reach it, and it is edible, but it would be a futile task to search nearby for more or to come to the same spot again a month later: you'll find nothing there.

So, since no deep idea seems to be lurking behind the scenes, let us just resort to a (reasonably) careful bookkeeping. We shall merely take the second derivative and show that it is positive.

**Identity 1:**
$$
\frac{d^2}{dx^2}\log(1\pm x^p)=\pm px^{p-2}\left[\frac p{(1\pm x^p)^2}-\frac 1{1\pm x^p}\right]\,.
$$
Now let $m=2n+1$. Put $u=x^m$ and consider 
$$
\Phi(x)=\log[(1+x^{2m-1})(1+x^{2m-3})]
$$
(those are just the first 2 factors in the full product).
We have
$$
\Phi''(x)=(2m-1)(2m-2-x^{2m-1})\frac {x^{2m-3}}{(1+x^{2m-1})^2}
+(2m-3)(2m-4-x^{2m-3})\frac {x^{2m-5}}{(1+x^{2m-3})^2}
\\
\ge
(2m-1)(2m-2-u)\frac {x^{2m-3}}{(1+u)^2}
+(2m-3)(2m-4-u)\frac {x^{2m-5}}{(1+u)^2}
$$
(we need that $m\ge 3$ here).

Think of this as our supply of positivity consisting of $(2m-1)(2m-2-u)$ silver units $\frac {x^{2m-3}}{(1+u)^2}$ and 
$(2m-3)(2m-4-u)$ golden units $\frac {x^{2m-5}}{(1+u)^2}$.

Now it is time to look at $\Psi(x)=\log\left[\frac{(1+x^{m-1})(1-x^m)}{(1+x^{m})(1-x^{m+1})}\right]$.
We would like to reduce the logarithm of every factor just to its leading term in the Taylor expansion. Let's take the expressions one by one and see what correction is needed for that.

**The power $x^{m-3}$ (the top left factor)**.
The actual contribution to $\Psi''(x)$ is $(m-1)x^{m-3}\left[\frac{m-1}{(1+x^{m-1})^2}-\frac 1{(1+x^{m-1})}\right]$. To raise it to $(m-1)(m-2)x^{2m-3}$, we need to add
$$
\frac{x^{m-3}}{(1+x^{m-1})^2}\left[(m-1)^2(2x^{m-1}+x^{2(m-1)})-(m-1)x^{m-1}(1+x^{m-1})\right]
\\ =\frac{x^{2m-4}}{(1+x^{m-1})^2}[(2m-3)(m-1)+(m-1)(m-2)x^{m-1}]
 \le (3m-5)(m-1)\frac{x^{2m-4}}{(1+x^{m-1})^2}\,.
$$
We want to dominate it by our units. Note that the golden ones have the power $2m-5$ while the silver ones have the power $2m-3$, so averaging one gold and one silver, we can dominate the power $2m-4$ in the estimates. Thus the cost of this upgrade is  $(3m-5)(m-1)$ units (the denominator we have is larger than $(1+u)^2$, so it works in our favor) 
Let us check that we have enough gold supply to pay, i.e., that  
$$
\frac 12 (3m-5)(m-1)\le (2m-3)(2m-4-u)\,.
$$
Even in the worst case scenario $u=1$, this reduces to $(3m-5)(m-1)\le (4m-6)(2m-5)$, which works for $m\ge 4$ (each factor is larger). Note however that we cannot afford paying in pure gold  for $m=4,5$ and have to mix at least partially there. However, this is the only case where the difference between gold and silver units matters. Everywhere else the power will be always in our favor.

**The power x^{m-2} (bottom left factor)**
Exactly the same argument applies to that factor (replacing $m-1$ by $m$, of course), except in this case we do not lose but, conversely, get a gain towards our positivity supply. The formula for the gain is pretty much the same as the loss formula before: our advantage is
$$
\frac{x^{2m-2}}{(1+x^{m})^2}[(2m-1)m+m(m-1)x^m]\ge \frac{x^{2m-2}}{(1+u)^2}m(m-1)(2+u)=:x^{2m-2}G\,.
$$
The last rough estimate (just losing $m$ for no apparent reason) looks a bit idiotic at the moment, but the only way we shall use this gain will be  to offset a certain loss term, so I just put it in the most convenient form for that future step.

**The factors with minuses**
Here we need another identity:
$$
\frac{d^2}{dx^2}\log(1- x^p)+p(p-1)x^{p-2}=-px^{2p-2}\left[\frac p{(1- x^p)^2}+\frac {p-1}{1- x^p}\right]\,.
$$
For the proof just take Identity 1 and play a bit with algebra.  If you want to run a quick mental sanity check, note that the denominators are of the same kind, the pole at $1$ has the same order and asymptotics, the asymptotics of the whole thing near $x=0$ agrees with that coming from the second term in the Taylor expansion of the corresponding logarithm, and that these  requirements together with the order of decay near $\infty$ in the complex plane allow one to recover the whole formula in a unique way. 
Applying it with $p=m$ and $p=m+1$, we see that the loss we need to compensate for is $x^{2m-2}$ times
$$
\left[\frac{m^2}{(1-x^m)^2}+\frac{m(m-1)}{1-x^m}\right]-x^2\left[\frac{(m+1)^2}{(1-x^{m+1})^2}+\frac{(m+1)m}{1-x^{m+1}}\right]
\\
=\left[\frac{m^2}{(1-x^m)^2}-x^2\left(\frac{(m+1)^2}{(1-x^{m+1})^2}+\frac{m+1}{1-x^{m+1}}\right)\right]
\\
+(m-1)\left[ \frac{m}{1-x^m}-x^2\frac{m+1}{1-x^{m+1}}  \right]=:L_1+L_2\,.
$$

Now comes the first interesting estimate: $G\ge L_2$. We shall use a simple inequality $\frac{m+1}{1-x^{m+1}}\ge \frac{m}{(1-x^m)}$ (I'll skip the proof for now because we shall prove a stronger result a few lines below) to estimate $L_2$ from above by $m(m-1)\frac{1-x^2}{1-x^m}$. Thus, it will suffice to show that
$$
(2+x^m)(1-x^m)=2-x^m-x^{2m}\ge (1+x^m)^2(1-x^2)\,. 
$$
Note that the LHS increases and the RHS decreases as $m$ goes up, so it will suffice to consider $m=3$, i.e., to check that (after division by $1-x$)
$$
(2+x^3)(1+x+x^2)-(1+x+2x^3+2x^4+x^6+x^7)\\ =1+x+2x^2-x^3-x^4+x^5-x^6-x^7\ge 0\,.
$$
However $2x^2\ge x^3+x^4$ and $x+x^5\ge x^6+x^7$. Thus, we can forget about $L_2$ and $G$ from now on and concentrate on $L_1$.

**To be continued...**