Show $\int_0^A \int_0^A \frac{1+(s-t)^2}{1+a^2(s-t)^2} e^{i (t-s) } dt ds\ge 0$ How to show that 
\begin{align}
\int_0^A  \int_0^A  \frac{1+(s-t)^2}{1+a^2(s-t)^2} e^{i  (t-s) } dt ds \ge 0
\end{align}
for $a>1$ for all $A>0$. 
Note the the above integral is real since the function
\begin{align}
f(t-s)= \frac{1+(s-t)^2}{1+a^2(s-t)^2}
\end{align}
is symmetric. 
So, in fact we have to show that
\begin{align}
\int_0^A  \int_0^A  \frac{1+(s-t)^2}{1+a^2(s-t)^2} cos(t-s) dt ds \ge 0
\end{align}
 A: By parity of the involved function, if $T$ is the region given by $0\leq y \leq x \leq A$,
$$ I=\iint_{(0,A)^2}\frac{1+(s-t)^2}{1+a^2(s-t)^2}\cos(t-s)\,ds\,dt = 2\iint_{T}\frac{1+(x-y)^2}{1+a^2 (x-y)^2}\cos(y-x)\,dx\,dy$$
and by a change of variable:
$$ I = 2 \int_{0}^{A}(A-u)\cos(u)\frac{1+u^2}{1+a^2 u^2}\,du. $$
Assuming $A\leq \pi$, the claim now follows from the second mean value theorem for integrals:
$$ I = 2 \int_{0}^{B}(A-u)\cos(u)\,du, \qquad B\in[0,A] $$
where the last constraint ensures:
$$ I = 2(1-\cos B)+2(A-B)\sin(B) \geq 0. $$
If $A>\pi$, we may re-group terms in the resulting integral according to the periodicity of the cosine function and apply (essentially) the same argument.

Alternative (and cleaner) solution. By the properties of the Fejér kernel, if $\omega(x)$ is an even function with a non-negative Fourier cosine transform, then
$$ \int_{0}^{A}(A-u) \omega(u)\geq 0. $$
On the other hand, if two even functions have non-negative Fourier cosine transforms, so does their product. $\cos(u)$ and $1$ trivially have non-negative Fourier cosine transforms, and the Fourier cosine transform of $\frac{1}{1+a^2 u^2}$ is given by $\sqrt{\frac{\pi}{2}} \frac{e^{-s/a}}{a}\geq 0$. It follows that by considering $\omega(u) = \cos(u)\frac{1+u^2}{1+a^2 u^2}$ the claim is trivial.
