Inspired by Prove that if $a+b =1$, then $\forall n \in \mathbb{N}, a^{(2b)^{n}} + b^{(2a)^{n}} \leq 1$. I propose :
Let $a,b>0$ such that $a+b=1$ then we have : $$\exp{\Big(\frac{2b(1-(2b)^n)}{n(1-2b)}\ln(a)}\Big)+\exp{\Big(\frac{2a(1-(2a)^n)}{n(1-2a)}\ln(b)}\Big)\leq1 $$
I try induction method but I have not a well evaluation of one of the term .
I try also to factorize with the formula :
$$e^{ix}+e^{iy}=2\cos(\frac{x-y}{2})e^{i\frac{x+y}{2}}$$ But I don't know if it's relevant here .
Maybe series expansion or Bernoulli's inequality can give something .
So if you can catch this problem ...
Thanks a lot for your time and patience .
Edit : The case $n=2$
Since the function (where $0\leq x \leq 1$ and $a+b=1$ and $a,b>0$):
$$f(x)=a^{4b^2(1-x)+2bx}+b^{4a^2(1-x)+2ax}$$
Is convex we have by Jensen's inequality :
$$f(0)+f(1)\geq 2f(0.5)$$
Or :
$$a^{4b^2}+b^{4a^2}+a^{2b}+b^{2a}\geq 2(a^{2b^2+b}+b^{2a^2+a})$$
But $2\geq a^{4b^2}+b^{4a^2}+a^{2b}+b^{2a}$ because If $a+b=1$ so $a^{4b^2}+b^{4a^2}\leq1$ And if $a,b>0$ and $a+b=1$ then $1\geq a^{2b}+b^{2a}$ https://eudml.org/doc/223938
And we are done .
For the general case maybe one can use Cauchy induction
