Playing with exponent I found this :
Let $a,b>0$ such that $a+b=1$ then we have : $$a^{\frac{1}{(2a)^n}}+b^{\frac{1}{(2b)^n}}\leq 1$$ Where $n\geq 1$ a natural number .
First of all I have tested until $n=30$ . The equality cases are $a=b=0.5$ or $a=0$ and $b=1$ .
I have tried to invoke the convexity but it does not work . For $n=1$ I can prove a part of the inequality using Bernoulli's inequality .I have tried to derivate but it's ugly .And finally I have make more progess using power series around $a=0.5$ .In fact we can approximate a part of the curve using parabolas .And after playing with exponent we see that we get an infinity of inequalities wich tend to $1$
So if you have a counter-example I accept or if you have an (the good) answer wich capt the truth of this.
Any helps is greatly appreciated.
Thanks a lot.
Ps:Sorry for the bad english .
Edit :
My proof :
Since we have a proof (see the link for Maximilian) of :
Let $a,b>0$ such that $a+b=1$ an $n\geq 1$ a natural number then we have : $$a^{(2b)^n}+b^{(2a)^n}\leq 1$$
Let prove this :
Let $a,b>0$ such that $a+b=1$ an $n\geq 1$ a natural number then we have : $$a^{\frac{1}{(2a)^n}}+b^{\frac{1}{(2b)^n}}\leq a^{(2b)^n}+b^{(2a)^n}\leq 1$$
The proof is very easy since we have :
$$a^{\frac{1}{(2a)^n}}\leq a^{(2b)^n}$$
Because it's equivalent to :
$$(4ab)^n\leq 1$$
Wich is true by Am-Gm and the condition $a+b=1$
Done !
My question: have you an alternative proof (using maybe convexity ?)
Ps:I think that Maximilian have seen that before me(it's a little bit mischievous ^^)