let $a,b>0$ such that $a+b=\frac{1}{2}$ then we have : $$a^{2b}+b^{2a}\leq 1$$
My try :
I have tried to apply Bernoulli's inequality :
$$f(a)=a^{2(\frac{1}{2}-a)}=(1+a-1)^{2(\frac{1}{2}-a)}\leq g(a)=1+(a-1)(2)(\frac{1}{2}-a)$$
On the other hand we have : $$f(\frac{1}{2}-a)\leq g(\frac{1}{2}-a)$$
Unfortunately we have :
$$g(\frac{1}{2}-a)+g(a)\geq 1$$
So I was wondering myself if we have some good bound using power series but I found nothing really interesting and usefull .
The equality case are $a=b=0.25$ or $a=0$ and $b=0.5$
Any helps is greatly appreciated .
Thanks in advance for all your contributions .
Update
Let $a\geq b>0$ such that $a+b=0.5$ then the function : $$(ax+b(1-x))^{2(a(1-x)+b(x))}=f(x)$$ Is $3$-concave or $f'''(x)\leq 0$ on $(0,1)$
So we can apply the Levinson's inequality the theorem B . It gives inequality like :
$$\frac{f(0)}{2}-f(0.125)\geq \frac{f(0.5)}{2}-f(0.375)$$
Maybe it's usefull.