Consider a "dented triangle" like in this picture:
(the angle $\alpha$ is larger than $\pi$). Compare it with an actual triangle $\triangle\bar{A}\bar{B}\bar{C}$ with sidelengths $a, b$ and $c = c' + c''$, with a point $\bar{X}$ on the side of length $c$ at distances $c'$ and $c''$ of the respective endpoints. I need to show that
$$
d(\bar{C}, \bar{X}) \geq d(C, X)
$$
($d$ denotes distance).
I do not only need this for a euclidean triangle, but also for hyperbolic and spherical triangles (in the last case maybe you need $a, b \leq \pi/2$)
Some background: This is the last (and I thought easiest) step in the proof of a "reverse" Toponogov comparison theorem which states that under an upper curvature bound on a Riemannian manifold triangles are "thinner" than the corresponding comparison triangles.
Ideas so far: By the corresponding cosine laws it is equivalent to show that the angle at $\bar{A}$ is larger than the one at $A$.
Also if both $a > c''$ and $b > c'$ then (at least in the euclidean case) it is easy to prove since one of the two angles at $X$ has to decrease when passing to the "straight triangle".
Notation : $i) \ [xy]$ is a line segment between $x$ and $y$
$ii)\ |A-B|$ is a distance between $A$ and $B$
Proof : Assume that $\Delta\ \overline{C}\overline{B}\overline{A}$ is a triangle s.t. $|\overline{C}-\overline{B}|=|C-B|,\ |\overline{C}-\overline{A}|=|C-A|,\ |\overline{B}-\overline{A}|=c'+c'' $
Consider a point $A'$ s.t. a line segment $[BA']$ contains $X$ and $|X-A'|=|X-A|$.
Since $\alpha >\pi$, then $\angle\ CXA > \angle \ CXA'$
Hence $|C-A|>|C-A'|\ \ast$. In a triangle $\Delta\ CBA'$, by $\ast$, then $\angle\ CBA' <\angle\ \overline{C} \overline{B}\overline{A}\ \ast\ast$
Hence from $\ast\ast$, we complete the proof.
Reference : Alexandrov Lemma in the book From Euclid to Alexandrov; a guided tour - Petrunin and Yashinski