The upper bound of a sum of exponential function

Question

Could someone help me to find the upper bound of the following function:

$f(x) = \sqrt{\sum_{n=i}^{N} e^{-\alpha_{i}\cdot x}}$,

where $x > 0$, the $i^{th}$ coefficient $\alpha_{i} > 0$.

I got one that $\sqrt{\sum_{n=i}^{N} e^{-\alpha_{i}\cdot x}} \leq \sum_{n=i}^{N}\sqrt{e^{-\alpha_{i}\cdot x}}$, but I feel it is too loose, I extremely expect for a more tight upper bound. Furthermore, it is better that the potential upper bound could be represented in the manner of being a function of $e^{-\alpha_{i}x}$

Answer 1

Let :

$$ \alpha_{\text{min}} = \min \lbrace \alpha_{i}, \; 1 \leq i \leq n \rbrace. $$

Then, for all $x > 0$ and for all $i \in \lbrace 1,\ldots,n \rbrace$, $e^{-\alpha_{i}x} \leq e^{-\alpha_{\text{min}}x}$.

Therefore :

$$ \forall x > 0, \; f(x) = \Big( \sum_{j=1}^{N} e^{-\alpha_{j}x} \Big)^{1/2} \leq \sqrt{N}e^{-\alpha_{\text{min}}x/2}. $$