Given $m \leq x_1,\ldots,x_n \leq M$, what is a good upper bound for $n\log\left(\frac{\sum_i e^{x_i}}{n}\right)$?
My approach: $e^{x_i} \leq 1 + \frac{e^M-1}{M}x$ for all $x \leq M$. Using this, we get $n\log\left(\frac{\sum_i e^{x_i}}{n}\right) \leq n\log\left(1+ \frac{e^M-1}{M}\frac{\sum_i x_i}{n}\right)$.
Now using $\log(1+x) \leq x$, we have $n\log\left(1+ \frac{e^M-1}{M}\frac{\sum_i x_i}{n}\right) \leq \frac{e^M - 1}{M} \sum_i x_i$.
I did not use the lower bound $m$ but in a similar way, I can get $n\log\left(\frac{\sum_i e^{x_i}}{n}\right) \leq nm + \frac{e^{M-m} - 1}{M-m} \sum_i (x_i-m)$.
Is there a better upper bound possible?
Maximizing $n\log\left(\frac{\sum_i e^{x_i}}{n}\right)$ is equivalent to maximizing $$ \sum_{i=1}^n e^{x_i}$$ under the constraints that
We notice that the function $x \mapsto e^x$ is a convex function (as $(e^x)'' = e^x >0$), by applying the Jensen's inequality to the two points $(m,M)$ with positive weights $\left(\frac{M-x_i}{M-m},\frac{x_i-m}{M-m} \right)$: $$\frac{M-x_i}{M-m}\cdot e^m +\frac{x_i-m}{M-m} \cdot e^M \ge \exp \left(\frac{M-x_i}{M-m}\cdot m + \frac{x_i-m}{M-m}\cdot M \right) = e^{x_i} \hspace{0.5cm} \forall i\in \{1,..,n \} \tag{1}$$ The equality of $(1)$ occurs if and only if $x_i = m$ or $x_i = M$. Make the sum of $(1)$ for $n = 1,..,n$, we have: $$\sum_{i=1}^n \left(\frac{M-x_i}{M-m}\cdot e^m +\frac{x_i-m}{M-m}\cdot e^M \right) \ge \sum_{i=1}^n e^{x_i}$$ $$\iff \color{red}{\sum_{i=1}^n e^{x_i} \le \frac{e^M - e^m}{M-m}\cdot C + n\cdot \sum_{i=1}^n \frac{Me^m-me^M}{M-m}} \tag{2}$$
The equality of $(2)$ occurs if all $x_i$ is equal to $M$ or $m$ only. i.e there exists a $k \in \{1,..,n\}$ such that: $m\cdot k + M \cdot (n-k) = C \iff k = \frac{Mn - C}{M-m} \in \mathbb{N}$ ( $k$ variables $x_i$ receive the value $m$, the other $(n-k)$ of $x_i$ receive the value $M$).
The tightest upper bound is then
$$n\log\left(\frac{\sum_i e^{x_i}}{n}\right) \le \color{red}{n\cdot \log\left( \frac{e^M - e^m}{M-m}\cdot \frac{C}{n}+ \sum_{i=1}^n \frac{Me^m-me^M}{M-m} \right)} $$
Remark: For the tightest lower bound, it suffices to apply the AM-GM inequality
$$n\log\left(\frac{\sum_i e^{x_i}}{n}\right) \ge n\cdot \log\left( \sqrt[n]{\prod_{i=1}^n e^{x_i}} \right) = n\cdot \log\left( e^{C/n} \right) = \color{red}{C} $$ The equality occurs if and only if $x_i = \frac{C}{n}$ for $i=1,..,n$.