In a convex quadrilateral $ABCD$ diagonals $AC$ and $BD$ intersect at point $S$. Denote with $P$ the center of circumcircle of triangle $ABS$. Denote with $Q$ the center of circumcircle of triangle $CDS$. Prove that:
$$4 PQ \geq AB + CD.$$
I was able to calculate the right side based on angle between diagonals and circle radii. I was also able to make some estimate of the distance between circle centers. But I could not make the match between two formulas.
A graph is drawn below, together with the explanation of the notations. To simplify the explanation, we only restrict to the most non-trivial case: $\angle B \geq \angle A$, $\angle C \geq \angle D$.
It is not hard to see that $\angle PSA = \frac{\theta + \angle A - \angle B}{2}$ and $\angle QSD = \frac{\theta + \angle D - \angle C}{2}$. Therefore
$$ \angle PSQ = \frac{2\theta + \angle A - \angle B + \angle D - \angle C}{2} + 180^\circ - \theta \geq 180^\circ - \frac{\angle B + \angle C}{2} \geq \theta. $$
By law of sines, $\overline{AB} + \overline{CD} = 2(r_1 + r_2)\sin\theta$.
By law of cosines,
$$ \begin{aligned} \overline{PQ} &\geq \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos\theta} = \sqrt{(r_1+r_2)^2 - 2(1+\cos\theta)r_1r_2}\\ &\geq\sqrt{(r_1+r_2)^2 - \frac{1}{2}(1+\cos\theta)(r_1+r_2)^2}\\ &= (r_1+r_2)\sqrt{\frac{1-\cos\theta}{2}} = (r_1+r_2)\sin\frac{\theta}{2} \geq \frac{1}{2}(r_1+r_2)\sin\theta. \end{aligned} $$