Prove that if $a+b =1$, then $\forall n \in \mathbb{N}, a^{(2b)^{n}} + b^{(2a)^{n}} \leq 1$.

This is only a comment, but too long. Here an example for a short consideration, $n\ge 1$:

If we look at $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}\enspace$ and $\enspace\displaystyle \frac{0.5-(1-x)^{(2x)^n}}{x-0.5}\enspace$ then we see numerically $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}>\frac{0.5-(1-x)^{(2x)^n}}{x-0.5}\enspace$ for $\enspace 0<x<0.5\enspace$ and because of the symmetrie $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}<\frac{0.5-(1-x)^{(2x)^n}}{x-0.5}\enspace$ for $\enspace 0.5<x<1$ . This is equivalent to your question.

We can find out, that the maximum of $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}\enspace$ is at $\enspace x_0\enspace$ (depending on $n$ ;

the function is increasing for $\enspace 0<x<x_0$) $\enspace$ and that the maximum of $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}\enspace$

is at $\enspace 1-x_0\enspace$ (the function is increasing for $\enspace 0<x<1-x_0$) $\enspace$ with $\enspace x_0<0.5<1-x_0\enspace$

$($e.g. $n:=2$ : $x_0\approx 0,472376195328360)\enspace$ which is interesting,

because at $\enspace x=0.5\enspace$ we have $\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}=\frac{0.5-(1-x)^{(2x)^n}}{x-0.5}\enspace$ for all $\enspace n\enspace$

and we’ve got a piece of circumstantial evidence that the inequalities are valid.

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Numerically we see that $\enspace\displaystyle x^{(2(1-x))^n}+(1-x)^{(2x)^n}\enspace$ has one minimum for $\enspace 0<x<0.5\enspace$ and the maximum is $1$ at $x=0$ and $x=0.5$ .

We have $\enspace\displaystyle\min(x^{(2(1-x))^{n_2}}+(1-x)^{(2x)^{n_2}})<\min(x^{(2(1-x))^{n_1}}+(1-x)^{(2x)^{n_1}})\enspace$ for $\enspace n_1<n_2$ which is a good hint that a proof for $n=1$ can include the proof for $n>1$ .

This special case $n=1$ can be found as $\,$ Proposition 5.2 $\,$ in the article of https://eudml.org/doc/223938 .

This is a WRONG proof.

Here is an attempt/sketch, definitely not a proof as there is one problematic spot:

$\cos^{2}(x) + \sin^{2}(x) = 1$, raise by $2^n$

$(\cos^{2}(x) + \sin^{2}(x))^{2^n} \ge \cos^{2}(x)^{2^n} + \sin^{2}(x)^{2^n}$

This is true since $n$ is a natural number and we are removing non-negative quantities from the left. We note that the right side contains two quantities which are less then $1$, yet by the inequality we still have:

$\cos^{2}(x)^{2^n} + \sin^{2}(x)^{2^n} \le 1$, now we create two more inequalities by raising by $\cos^{2n}(x)$ and by $\sin^{2n}(x)$

(The following two inequalities I have no proof for): By similar reasoning, we get $\cos^{2}(x)^{2^n\cos^{2n}(x)} + \sin^{2}(x)^{2^n\cos^{2n}(x)} \le 1$, and

$\cos^{2}(x)^{2^n\sin^{2n}(x)} + \sin^{2}(x)^{2^n\sin^{2n}(x)} \le 1$

If we add these two inequalities and rearrange them we get

$\cos^{2}(x)^{2^n\sin^{2n}(x)} + \sin^{2}(x)^{2^n\cos^{2n}(x)} \le 2 - (\cos^{2}(x)^{2^n\cos^{2n}(x)} + \sin^{2}(x)^{2^n\sin^{2n}(x)} )$

We note that the functions on the right are of the form $f(x)=x^{2^nx^{2n}}=x^{(2x^2)^n}$ with symmetry $x \to 1-x$. By looking at the derivative (or at the graph) we see that the function in the interval $[0;1]$ attains a minimum in the interval $[\frac{1}{\sqrt e};1]$, but in the interval $[0;1/2 + \epsilon]$ approaches 1 as $n \to \infty$. In other words $f(x) + f(1-x) \ge 1, x \in [0,1]$ by inspection.

Hence: $\cos^{2}(x)^{2^n\sin^{2n}(x)} + \sin^{2}(x)^{2^n\cos^{2n}(x)} \le 1$

Now set $ a = \cos^2(x), b=\sin^2(x)$ to get:

$(a)^{2^nb^{n}} + (b)^{2^na^{n}} \le 1$ or

$a^{(2b)^n} + b^{(2a)^n} \le 1$ as desired.}