I want to solve the following optimization problem 

\begin{align}
\inf_{ X: |X| \le a \text{ a.s.}}   E \left[ \frac{1}{1+(X-X^\prime)^2} \right]
\end{align}
where $X^\prime$ is an independent copy of $X$ and $a>0$ is some constant.  

 How would one approach such a problem? Is the solution easy to find? 


At some point I thought that the optimal distribution is given by $X=\{-a,a\}$ equally likely.  In which case, the solution is given by
\begin{align}
\inf_{ X: |X| \le a \text{ a.s.}}   E \left[ \frac{1}{1+(X-X^\prime)^2} \right] \le \frac{1}{2}\frac{1}{1+4a^2}+\frac{1}{2}. 
\end{align}
However, I don't have any supporting arguments for this. 


The following might be useful. 
Note that by Jensens' inequality 

\begin{align}
E \left[ \frac{1}{1+(X-X^\prime)^2} \right] \ge  \frac{1}{1+E[(X-X^\prime)^2]} =\frac{1}{1+2Var(X)}.
\end{align}

Therefore,

\begin{align}
\inf_{ X: |X| \le a \text{ a.s.}}   E \left[ \frac{1}{1+(X-X^\prime)^2} \right]  \ge  \frac{1}{1+2 \sup_{ X: |X| \le a \text{ a.s.}} Var(X)}=\frac{1}{1+2a^2},
\end{align}

where in the last optimization step we used
\begin{align}
Var(X) \le E[X^2] \le a^2,
\end{align}
which is achievale with $X=\{-a,a\}$ equally likely.