Is $g(u)= \frac{E [ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} ] }{E [ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} ]}$ decreasing in $u$ Let $X$ be a positive random variable, let us define a function
\begin{align}
g(u,a)= \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} \right]}.
\end{align}
Question: Can we show that the above integral is monotonically decreasing in $u$  ( for $u>0$ ) for all $a > 1$.
Note that $X$ here represents the variance of standard normal. That is we consider the variance to be a random variable. 
I can show that $g(u,a)$ is bounded by $1$ and continuous but can not establish that it is decreasing.  Also, note that the function $g(u,a)$ is symmetric around $u=0$.
What I tried: 
I was able to show that  for $p,q\ge 1$ and $\frac{1}{q}+\frac{1}{p}=1$ and $a^2 \ge \frac{1}{p}$ we have
\begin{align}
g(u,a) \le  \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
where $\beta=\sqrt{\frac{q(a^2-\frac{1}{p})}{a^2}}$.
Proof:
By using Holder's inequality
\begin{align}
E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] &=E \left[ \frac{1}{\sqrt{X}} e^{-\frac{ (a^2-\frac{1}{p})u^2}{2X}} e^{-\frac{ \frac{1}{p}u^2}{2X}}  \right] \\
&\le E^\frac{1}{q} \left[ \frac{1}{\sqrt{X}} e^{-\frac{ q(a^2-\frac{1}{p})u^2}{2X}}   \right]  E^\frac{1}{p} \left[ \frac{1}{\sqrt{X}}  e^{-\frac{ u^2}{2X}}  \right].
\end{align}
Therefore,
\begin{align}
g(u,a) \le \left( \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{q(a^2-\frac{1}{p})u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} \right]} \right) ^\frac{1}{q}
&=  \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
and 
\begin{align}
g(u,a) \le  \left( g( \beta \cdot u, a ) \right)^{\frac{1}{q}},
\end{align}
where $\beta=\sqrt{\frac{q(a^2-\frac{1}{p})}{a^2}}$.
Thank you. Looking forward to seeing your approaches.

This is false in general.
Change the variables as in my comment: $1/(2X) \to X$, $a^2 \to a$, $u^2 \to u$. Then the problem is to show that 
$$
f(u) = \frac{\mathsf{E}[\sqrt{X}e^{-auX}]}{\mathsf{E}[\sqrt{X}e^{-uX}]}.
$$
decreases. 
Set $a=1.1$ and let $X=1$ or $100$ with probability $1/2$. Then
$$
f(u) = \frac{e^{-1.1u} +10e^{-110u}}{e^{-u} +10e^{-100u}}.
$$
However, it is not decreasing for small values of $u$: Wolfram Alpha

