<s>[INST] Show for $0 < A < B, \int_{-\infty}^\infty \frac{A^2 - A + B^2 - B + (A + B - 2AB)\cos(v) + v^2}{(A^2-2AB\cos(v)+B^2+v^2)^2}dv > 0$. The integral in question doesn't have a closed form solution as far as I know, but it does converge. The denominator is clearly positive, and the numerator of the integrand is always positive, except in a very narrow situation, which I will describe:
We have that $A^2 + B^2 - 2AB\cos(v) \geq (B-A)^2 > 0$. Furthermore, we have $\cos(v) \geq 1 - \frac{v^2}{2}$ by the Taylor formula, so we also have $(A + B)(\cos(v) - 1) + v^2 \geq (A + B)(- \frac{v^2}{2}) + v^2 = v^2(1 - \frac{A+B}{2})$. So only in the limited scenario where $\frac{A+B}{2} > 1$ could the integrand be negative. Now I'm stuck trying to prove this would not be enough to force the entire integral to be negative.
This inequality is part of a conjecture I have, so there is a chance it could be wrong!
 [/INST] If $A + B - 2AB \le 0$, then
$$\mathrm{NUM} \ge A^2 - A + B^2 - B + (A + B - 2AB) = (A - B)^2 > 0.$$
If $A + B - 2AB > 0$ and $A + B \ge 2$, then
$$\mathrm{NUM}
\ge A^2 - A + B^2 - B - (A + B - 2AB) = (A + B)(A + B - 2) \ge 0 .$$
If $A + B - 2AB > 0$ and $A + B < 2$, using $\cos v \ge 1 - v^2/2$, then
\begin{align*}
 \mathrm{NUM} &\ge A^2 - A + B^2 - B + (A + B - 2AB)(1 - v^2/2) + v^2\\
 &= \frac12(2AB + 2 - A - B)v^2 + (A - B)^2\\
 &\ge 0.
\end{align*}
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