I didn't find it on Aops so let me propose it. It's the case $n=4$ of Prove that $a^{4/a} + b^{4/b} + c^{4/c} \ge 3$
Problem:
Let $a,b,c,d>0$ such that $a+b+c+d=4$ then it seems we have:
$$a^{\frac{4}{a}}+b^{\frac{4}{b}}+c^{\frac{4}{c}}+d^{\frac{4}{d}}\geq 4$$
In the link below I propose a sketch of proof using the excellent idea of Professor Vasile Cirtoaje.
My arguments go as follows:
@Andreas has already shown that $f\left(x\right)=x^{\frac{4}{x}}$ is convex for $x\in(0,1]$ .So using WLCF-Theorem with an obvious statement we need to show for $1\leq x\leq 4$ :
$$\frac{1}{4}f\left(x\right)+\frac{3}{4}f\left(\frac{4-x}{3}\right)\geq 1\tag{I}$$
See the reference for the WLCF-Theorem. Using Desmos I'm pretty sure $(I)$ works. On the other hand I think the case $n=5$ can fail.
How to (dis)prove it properly ?(solved I don't expose the proof here because it's not so pretty)
I ask again with a second question:
With the same constraint as above how to show :
$$a^{\frac{4+\frac{4}{5}}{a+\frac{4}{5}}}+b^{\frac{4+\frac{4}{5}}{b+\frac{4}{5}}}+c^{\frac{4+\frac{4}{5}}{c+\frac{4}{5}}}+d^{\frac{4+\frac{4}{5}}{d+\frac{4}{5}}}\geq 4$$
?
Reference :
Cirtoaje, V., Baiesu, A. An extension of Jensen's discrete inequality to half convex functions. J Inequal Appl 2011, 101 (2011). https://doi.org/10.1186/1029-242X-2011-101
