Assume you have a triangle with vertices $A,B,C\in\mathbb H^2$ in the hyperbolic plane and the hyperbolic distances are $a=d(B,C), b=d(A,C), c=d(A,B)$.
Now pick a `comparison triangle' $A',B',C'\in\mathbb R^2$ with the same side-lengths $a,b,c$ as our hyperbolic triangle.
Is it true that each angle $\alpha=\angle_{CAB},\beta=\angle_{ABC},\gamma=\angle_{BCA}$ in the hyperbolic triangle is bounded above by the corresponding angle in the Euclidean triangle, i.e. $\alpha\leq\alpha',\beta\leq\beta',\gamma\leq\gamma'$?
I tried the approach of comparing the cosines of the angles via the respective laws of cosines: $$\cos\alpha'=\frac{a^2+b^2-c^2}{2ab}\overset?\leq\frac{\cosh a\cosh b-\cosh c}{\sinh a\sinh b}=\cos\alpha,$$ however, I was not able to prove this inequality yet.
At least one of $\alpha, \beta, \gamma$ is smaller because $\alpha+ \beta+ \gamma < \pi .$ Then we use $$ \frac{ \sin \alpha}{\sinh a} = \frac{ \sin \beta}{\sinh b} = \frac{ \sin \gamma}{\sinh c} $$ and the fact that $\sinh x > x$ for $x > 0$