Let $r=2n+2$ where $n\in\mathbb{N}$
When $r\le 10$, is it possible to choose $\nu >0,\mu >\nu $ such that
$$\max(\frac{1}{6},\frac{1}{8}+\frac{3}{8(r-1)})<\nu<\frac{1}{4}+\frac{1}{2(r-1)}$$
and $$\max(\frac{1}{6},\frac{1}{12}+\frac{\nu}{3})<\mu<\frac{1}{3}+\frac{1}{2(r-1)}-\frac{2\nu}{3}$$
Thanks in advance
It is easy to check when $r\le 10 $ then $\tfrac{1}{6}\le\tfrac{1}{8}+\tfrac{3}{8(r-1)}$.
If $r=2$ then there is no such $\nu$. Indeed, we have $\tfrac 18+\frac 3{8(r-1)}=\frac 12<\nu$. On the other hand, $\nu<\mu<\frac 13+\frac{1}{2}-\frac {2\nu}{3}$ which implies $\nu<\frac 12$.
If $r\ge 4$ then we have to pick $\tfrac{1}{8}+\tfrac{3}{8(r-1)}<\nu<\tfrac {1}{2}+\tfrac{1}{2(r-1)}$. We have to pick $\mu>\nu$. Since $\nu>\tfrac{1}{8}+\tfrac{3}{8(r-1)}\ge\tfrac 16$ and $\nu>\tfrac 18$, this also assures conditions $\mu>\tfrac 16$ and $\mu>\tfrac 1{12}+\tfrac{\nu}{3}$. So to assure a possibility to chose $\mu$ it remains to satisfy a condition $\nu<\tfrac{1}{3}+\tfrac 1{2(r-1)}-\tfrac{2\nu}{3}$, that is $\nu<\tfrac{3}{5}+\tfrac 3{10(r-1)}$. This can be achieved iff $\tfrac{1}{8}+\tfrac{3}{8(r-1)}<\tfrac{3}{5}+\tfrac 3{10(r-1)}$, that is $3<19(r-1)$.