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.

**To be continued...**