Generalization of consequence of law of cosine There is the immediate consequence of the law of cosine stating that when fixing two sidelengths of a triangle and increasing the third, the vertex angle opposite of the third side increases as well. Moreover, this increase of the third side causes the adjacent angles to decrease. I feel like the following generalization of this should hold:
Let $a,b,c$ and $a',b',c$ be the sidelengths of two triangles and let $\gamma$ and $\gamma'$ be the vertex angles opposite of the side $c$ (which has the same length in both triangles). If $a+b \leq a'+b'$, then $\gamma \geq \gamma'$.
Simply put, if the sum of the lengths of the adjacent sides of an angle increases (and the opposite length is kept fixed) then this angle decreases.
Ideally, this should hold not only in the plane but in any space where the law of cosine holds, specifically I am thinking of the sphere and hyperbolic space.

I am having trouble proving this but my gut tells me this has to be true (I did not find anything online, although admittedly I am unsure how to search for this). Intuitively, since $c \leq a+b \leq a'+b'$, the detour through $A$ is "closer" to the shortest connection from $C$ to $D$ than the detour through $B$, so $A$ is closer to the segment $CD$ than $B$ is. This should somehow imply that the angle at $A$ is larger than the angle at $B$ (the angle increases when approaching a segment, with the extreme case of an angle of $\pi$ if the point is on the segment).
I am grateful for any further information!
 A: From triangle-exterior-angle-theorem \begin{align}\gamma =\gamma''+\alpha'=(\gamma'+\alpha)+\alpha' \,(\alpha\geq0,\alpha'\geq0)\end{align}
\begin{align}\gamma  > \gamma'\, (\alpha \neq0)\end{align}
\begin{align}\gamma = \gamma' \,(\alpha=\alpha'=0)\end{align}

A: 
Given $\triangle ABC$ with "vertex" $C$ ...

*

*As a consequence of the Inscribed Angle Theorem, the set of points $C'$ (on the same side of $\overline{AB}$ as $C$) such that $\angle AC'B>\angle ACB$ is bounded by the circumcircular arc $ACB$.

*On the other hand, the set of points $C'$ such that $|AC'|+|BC'|<|AC|+|BC|$ is bounded by the ellipse through $C$ with foci $A$ and $B$.

The circular "larger angle" region never fully covers the elliptical "smaller sum-of-sides" region, so the conjecture is false: the smaller sum-of-sides does not imply a larger angle. $\square$
Note: When $C$ happens to be an endpoint of the minor axis of the ellipse (that is, when $\triangle ABC$ is isosceles with vertex $C$), the "smaller sum-of-sides" region covers the "larger angle" region; so, in this case, a converse of the conjecture holds: a larger angle implies a smaller sum-of-sides. But, generally, how the angles compare is independent of how the sums-of-sides compare.
