Let $a,b>0$ such that $a+b=1$ and $n\geq 2$ a natural number then we have :
$$\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{a^n2^{n-1}}+\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{b^n2^{n-1}}\leq 1 $$ We have here Prove that if $a+b =1$, then $\forall n \in \mathbb{N}, a^{(2b)^{n}} + b^{(2a)^{n}} \leq 1$. where users discuss about an upper bound I remark that there is an interval $I=[\alpha,1]$ such that :
$$\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{a^n2^{n-1}}\leq b^{(2a)^{n}}$$ And : $$\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{b^n2^{n-1}}\leq a^{(2b)^{n}}$$
Because it's equivalent to : $$\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)\leq a^{2}$$
Or :
$$\frac{a}{b}\leq a^{2a}+b^{2b}$$
By convexity of $x^{2x}$ and Jensen's inequality we have :
$$1\leq a^{2a}+b^{2b}$$
Remains to show :
$$\frac{a}{b}\leq 1$$ wich proves one of the two inequalities .
For small $n$ we can use Bernoulli's inequality .
Another try :
Let $a,b>0$ such that $a+b=1$ and $n\geq 2$ a natural number then we have :
$$\sin^2\Bigg(\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{a^n2^{n-1}}\Bigg)+\sin^2\Bigg(\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{b^n2^{n-1}}\Bigg)\leq 1 $$
A bit of algebra shows that it's equivalent to the initial inequality .
Moreover :
$$\sin^2\Bigg(\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{a^n2^{n-1}}\Bigg)+\sin^2\Bigg(\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{b^n2^{n-1}}\Bigg)\leq \sin^2\Bigg(\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{a^n2^{n-1}}\Bigg)+\cos^2\Bigg(\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{a^n2^{n-1}}\Bigg) $$
Or : $$\sin^2\Bigg(\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{b^n2^{n-1}}\Bigg)\leq \cos^2\Bigg(\Bigg(\frac{ab}{a^{2a}+b^{2b}}\Bigg)^{a^n2^{n-1}}\Bigg) $$
Wich have the form :
$$\sin^2\Big(\beta^{b^n2^{n-1}}\Big)\leq \cos^2\Big(\beta^{a^n2^{n-1}}\Big)$$
Maybe for this last inequality we can use Taylor series and compare the coefficients .
Any helps is appreciated .
Thanks a lot for all your contributions .
