I was wondering myself about an inequality found on a maths website (but unfortunately now deleted ):
Let $a,b,c>0$ such that $a+b+c=1$ then we have : $$3(\frac{1}{3})^{(\frac{1}{3})!}\geq a^{b!}+b^{c!}+c^{a!}\geq 1$$
I have spent a lot of time on this inequality and first my idea was to make symmetry because we have : $$2(\frac{a+b}{2})^{(\frac{a+b}{2})!}\geq a^{b!}+b^{a!}$$ To be true this last inequality needs the condition of the beginning ($a+b+c=1$). So it delete the cyclicity . Secondly my idea was to use the Jensen's inequality but the function $x^{x!}$ is not concave between 0 and 1. Maybe someone can achieve this. Thanks a lot for your time.