Upper bound for sum of exponentials

Question

I want to show that for any vector $a\in \mathbb{R}^n$ with $\sum\limits_{i=1}^n a_i = 0$ and $m=\textrm{max}_i |a_i|$, it holds $$\frac{1}{n} \sum\limits_{i=1}^n e^{a_i} \leq \cosh (m)\, .$$ I really appreciate any ideas.

Answer 1

Let $n$ is even and $\max\limits_{i}a_i=m$.

Since $e^x$ is a convex function, by Karamata we obtain: $$\frac{1}{n}\sum_{i=1}^ne^{a_i}\leq\frac{1}{n}\left(\frac{n}{2}e^{-m}+\frac{n}{2}e^m\right)=\frac{e^m+e^{-m}}{2}=\cosh{m}.$$ The case $n$ is odd for you.