When is this conclusion true?

Question

Assume $A,B,C,D$ are arcs in the a circuit of radius $R=1$, all $A,B,C,D \le \pi$. What is/are the main constraint(s) for the following conclusion to be correct? $$ A+B < C+D \implies \sin(A/2)+\sin(B/2) < \sin(C/2)+\sin(D/2). $$ Your help is appreciated.

Answer 1

This is an elaborate comment ...

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I'll restate the problem this way, incorporating the fact that the chord lengths are proportional to the sines of the half-angles:

Let $P$, $Q$, $R$, $S$ be taken in order around a circle, such that $\stackrel{\frown}{PQ}$, $\stackrel{\frown}{QR}$, $\stackrel{\frown}{RS}$, $\stackrel{\frown}{SP}$ each has length no greater than $\pi$ (but the sum of those lengths is $2\pi$). Determine the conditions for which we can state: $$|\stackrel{\frown}{PQ}|+|\stackrel{\frown}{QR}| < |\stackrel{\frown}{RS}|+|\stackrel{\frown}{SP}| \quad\implies\quad |\overline{PQ}|+|\overline{QR}| < |\overline{RS}| + |\overline{SP}| \tag{$\star$}$$ That is, $$\text{When does the smaller arc-sum correspond to the smaller chord-sum?}$$

To see that the answer isn't "always", consider this situation:

In $\bigcirc O$, we take $O$ outside $\triangle PQR$, so that $\stackrel{\frown}{PQR}$ is a minor arc, and we contemplate positions for point $S$ on the major arc $\stackrel{\frown}{PR}$. Certainly, for any such $S$, the arc-sum inequality in $(\star)$ holds. Writing $P^\prime$ and $R^\prime$ for the points diametrically opposite $P$ and $R$, we see that $S$ must be common to semicircles $\stackrel{\frown}{PP^\prime}$ and $\stackrel{\frown}{RR^\prime}$ in order to satisfy the condition $|\stackrel{\frown}{PS}|\leq \pi$ and $|\stackrel{\frown}{RS}|\leq \pi$; that is, $S$ must lie on minor arc $\stackrel{\frown}{P^\prime R^\prime}$.

The diagram shows an ellipse with foci $P$ and $R$, passing through $Q$. By definition, this is the set of all points $X$ such that $$|\overline{PQ}|+|\overline{QR}| = |\overline{PX}|+|\overline{XR}| \tag{$1$}$$ Moreover, points $X$ inside the ellipse are such that the "$=$" in $(1)$ becomes "$>$".

To satisfy the chord-sum inequality in $(\star)$, point $S$ must lie on $\stackrel{\frown}{P^\prime R^\prime}$ but outside the ellipse.

(Conceivably, the ellipse could cover the circle, leaving no possible $S$. The reader is invited to investigate whether this ever actually happens.)