It seems the following. 

Put $\gamma=\gamma(X,\lambda)$ and 
$S=\sum_{m \in I} e^{-\gamma x_m}$.
The equality 
$$ \sum_{m \in I} (\lambda-x_m) e^{-\gamma x_m}=0$$
implies 
$$\lambda S=\sum_{m \in I} \lambda e^{-\gamma x_m}= 
\sum_{m \in I\setminus{1}} x_m e^{-\gamma x_m}\ge 
\sum_{m \in I\setminus{1}} s e^{-\gamma x_m}=s(S-1).$$

It transforms to $S\le \frac s{s-\lambda}=(1-\lambda/s)^{-1}$.