I am looking for proofs of or counterexamples to the following assumptions: 


if the corners of a triangle are chosen from a compact subset $\mathcal{S}$ of the Euclidean plane then all three corners of the following triangles are elements of the convex hull $CH(\mathcal{S})$ of $\mathcal{S}$ 


* the triangles with corners $A,B,C\ \in\ \mathcal{S}$,$\ $ that have the largest incircle
* the triangles with corners $A,B,C\ \in\ \mathcal{S}$,$\ $ that maximize $a+b-c$, where $\left(c:=\|B-A\|\right) \ge \left(b:=\|A-C\|\right)\ \ge\ \left(a:=\|C-B\|\right)$


My motivation is to identify "fat" triangles, for which it is guaranteed, that their corners are elements of the convex hull of some finite planar point sets. 



Extremality of Triangles with Corners from Planar Convex Hulls
1) No. Consider the square $ABCD$. Then the triangle $ABC$ does not have the largest incircle. Indeed, it has smaller incircle than $ABM$, where $M$ is the midpoint of $CD$. This follows from the formula $r(\Delta)=S(\Delta)/p(\Delta)$, where $r,S,p$ denote the inradius, area and semi-perimeter of a triangle $\Delta$. Perimeter of $\triangle ABM$ is smaller than that of $\triangle ABC$, it is seen from considering the point $K$ symmetric to $B$ against $C$: $$2p(ABM)=AB+AM+BM=AB+AM+MK=AB+AK<AB+AC+CK=2p(ABC).$$


2) No, the same counterexample works.


