Sharp Exponential Inequality of $\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}}$ I am trying to find a sharp inequality of $\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}}$ s.t. we can write: $$\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}} \leq \frac{e^{-ax}}{x}f(an) .$$ Here, $a,x>0, n \in \mathbb{N}. $ Any particular technique to answer these kinds of questions will be very helpful. Thanks.

The RHS approaches infinity as x approaches 0. Also, if $f(an)$ has the value 1, then the RHS is always greater than LHS. The derivative of $\frac{e^{−ax}}{x}$ is negative for all $a,x>0$, and that of the LHS is also negative.
Dividing the LHS by $\frac{e^{−ax}}{x}$ and taking the limit as $x\rightarrow \infty$ would give $f(an)$. We get
\begin{align}\frac{1}{\sqrt{1+\frac{n^2}{x^2}}}\exp{(ax(1-\sqrt{1+\frac{n^2}{x^2}}))}\le f(an)\end{align}
Binomial expansion of the square roots and dismissal of terms of order higher than $\frac{n^2}{x^2}$:
\begin{align}\frac{e^{-\frac{an^2}{2x}}}{1+\frac{n^2}{2x^2}}\le f(an)\end{align}
Thus, \begin{align}1\le f(an)\end{align}
EDIT
A suggestion in the comments states an easy way to do this
\begin{align}\frac{1}{\sqrt{1+\frac{n^2}{x^2}}}\exp{(-\frac{an^2}{\sqrt{n^2+x^2}+x})}\le f(an)\end{align}
which, upon taking the limit as $x\rightarrow \infty$, gives the same result.
