Coarse triangle minsize is bounded by square root of area

Finite nonempty sets have attained distance minima. At each disk vertex $v$ define $s(v)=d(f(v),A_1)$ and $t(v)=d(f(v),A_2)$, and extend the pair affinely over each triangle. The triangle inequality gives $\mathrm{\mid }d(x,A_i)-d(y,A_i)\mathrm{\mid }\le d(x,y)$ by using a nearest point for each of $x,y$ in turn. Consequently both coordinate differences on an edge are at most $r$. On the first arc $s=0$ throughout each edge and $t\ge 0$; on the second $t=0$ and $s\ge 0$; their common endpoint maps to the origin.

At any vertex of the third arc, choose nearest $a_1\in A_1,a_2\in A_2$ to its image $a_3$. If both coordinates were less than $m\mathrm{/}2$, then $d(a_1,a_3)<m\mathrm{/}2$, $d(a_2,a_3)<m\mathrm{/}2$, and $d(a_1,a_2)\le d(a_1,a_3)+d(a_3,a_2)<m$. All three pair distances would be less than $m$, contradicting its finite minimum definition. Hence $\max (s,t)\ge m\mathrm{/}2$ at each third-arc vertex.

On an edge of that arc start at either endpoint. A coordinate which is at least $m\mathrm{/}2$ there decreases by at most $r$ along the affine segment. Thus $\max (s,t)\ge h:=m\mathrm{/}2-r$ everywhere on the third arc. Its axis endpoints also have their nonzero coordinate at least $m\mathrm{/}2\ge h$. If $h>0$, all hypotheses of the crossing lemma now hold, so $h^2\le N{r}^2$. This deduction retains the loss of $r$ between vertices; the vertex barrier alone would not suffice.

Put $q=\sqrt{N}\ge 0$. Then $(rq{)}^2=N{r}^2$. For nonnegative numbers squaring preserves order, because $b^2-a^2=(b-a)(b+a)$. Therefore $h>0$ and $h^2\le (rq{)}^2$ imply $h\le rq$, so $m\le 2r\sqrt{N}+2r$. If $h\le 0$ then $m\le 2r$, which implies the same bound. This also treats $r=0$ and $N=0$ whenever such data are supplied.

Choose a minimizing vertex triple $(a_1,a_2,a_3)$. By the Hausdorff hypothesis, for every $\eta >0$ there are points $b_i$ on the respective continuous sides with $d(a_i,b_i)\le e+\eta$. Thus $d(b_i,b_j)\le d(a_i,a_j)+2e+2\eta \le m+2e+2\eta$. Taking the infimum over side triples and then letting $\eta \downarrow 0$ gives continuous minsize at most $m+2e$. Combining with step 4.1 proves the assertion. If the side sets are compact the distances are attained, but this limiting argument does not require attainment.