We are given $t>0$ and $a>0$. Compute $$\operatorname{inf}_{x>0}\; e^{-xt}(1-2a^2x)^{-1/2}$$
So if I compute even the minimum by considering the derivatives, we see that this is obtained for $x=\frac{t-a^2}{2a^2t}$, and therefore the infimum is given by $\frac{\sqrt t}{a^2} \exp\left(\frac{1}{2}-\frac{t}{2a^2}\right)$.
But what do I do if $t<a^2?$
Hint: The domain of the function is $(0,\frac 1{2a^{2}})$. If $f'(x)$ vanishes at only one point in the real line and you are excluding that point from the domain (as in this case when $t <a^{2}$) then $f'$ is always positive or always negative and $f$ is montonic in its domain. The infimum of such a function is the least of the values or the limiting values at the end points. Hence you only have to find $\lim_{x \to 0+} f(x)$ and $\lim_{x \to c-} f(x)$ where $c=\frac 1 {2a^{2}}$. The smaller of these two numbers is the infimum of $f$.