Let $X=(x_n)_{n \in I}$ be a sequence where $I=\{1,2,...,n\}$ with $n \ge 2$ and possibly $n =\infty$ is an index set with $x_1=0$ and $x_n>x_{n-1}$ for $n,n-1 \in I.$ We require that the sequence $X$ satisfies $$ \sum_{n \in I} e^{-\gamma x_n}<\infty \text{ for all }\gamma>0.$$

We then fix $x_2>0$ and define $\gamma(X,\lambda)>0$ for $\lambda \in (0,x_2)$ by the condition $$ \sum_{n \in I} (\lambda-x_n) e^{-\gamma(X,\lambda) x_n}=0.$$ Based on some examples, I then claim that for $\lambda \in (0,x_2)$ $$\max_{X;x_2 \text{ is fixed}} \sum_{n \in I} e^{-\gamma(X,\lambda) x_n} $$ is attained for the sequence $X=(0,x_2)$, i.e. $I=\{1,2\}$ such that

$$\max_{X;x_2 \text{ is fixed}} \sum_{n \in I} e^{-\gamma(X,\lambda) x_n} = 1 + e^{-\gamma((0,x_2),\lambda)x_2}.$$