A Euclidean right triangle has minsize proportional to its scale

Parameterize the two axis sides by ${\gamma }_1(u)=(u,0)$ and ${\gamma }_2(u)=(0,u)$ for $0\le u\le s$. Their pairwise parameter distances are $\sqrt{(u-v{)}^2}=\mathrm{\mid }u-v\mathrm{\mid }$. Parameterize the third side by ${\gamma }_3(u)=(u\mathrm{/}\sqrt{2},s-u\mathrm{/}\sqrt{2})$ for $0\le u\le s\sqrt{2}$. Its squared distance between parameters $u,v$ is $2((u-v)\mathrm{/}\sqrt{2}{)}^2=(u-v{)}^2$, so it too has distance $\mathrm{\mid }u-v\mathrm{\mid }$. These are isometric segments with exactly the displayed endpoints. Their lengths sum to $(2+\sqrt{2})s$.

Write an arbitrary side triple as $p=(u,0)$, $q=(0,v)$, $r=(t,s-t)$ with $u,v,t\in [0,s]$. Then $d_2(p,r)\ge s-t$ and $d_2(q,r)\ge t$, since their squared distances include respectively $(s-t{)}^2$ and $t^2$ and all other summands are nonnegative. Therefore its diameter is at least $\max (t,s-t)\ge s\mathrm{/}2$. This holds for every triple, so $\mu (s)\ge s\mathrm{/}2$.

Take $p=q=(0,0)$ and $r=(s,0)$, which belong respectively to the two axis sides and the third side. Their three pair distances are $0,s,s$, so the diameter is $s$ and $\mu (s)\le s$. In particular at $s=1$ the computed bounds are $1\mathrm{/}2\le \mu (1)\le 1$.

Multiplication of both coordinates by $s$ bijects all triples for scale one with all triples for scale $s$, with inverse division by $s$. The distance formula gives $d_2(sx,sy)=s{d}_2(x,y)$ because $s>0$. Hence the set of admissible diameters is exactly $s$ times the scale-one set. Multiplication by a positive scalar commutes with its infimum: all scaled values are at least $s\mu (1)$, and a value less than $\mu (1)+\epsilon \mathrm{/}s$ scales to less than $s\mu (1)+\epsilon$. Thus $\mu (s)=s\mu (1)$, with a positive coefficient by step 2.2.