Upper bound for sum of exponentials

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.

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.
• @Philmore I am ready to explain. By the way, I think it's better to wright $\max\limits_{i}|a_i|=m$. I am ready to fix my post if you want. – Michael Rozenberg May 26 '17 at 16:14
• @Michael_Rozenberg Your post is great. Obviously one does not have to look at two different cases by looking at the double-sized vector with two copies of every element of $a$. I would have had a really hard time proving above inequality without your suggestion of using Karamata. – Philmore May 26 '17 at 20:05