Let $\mathbf{D}$ be the unit disk, is $$\inf_{\begin{array}{c} v_{1},v_{2},v_{3},v_{4}\in\mathbf{D},\\ v_{0}\in\mbox{convexhull}\left(v_{1},v_{2},v_{3},v_{4}\right) \end{array}}\max_{0\le i,j,k\le4}\frac{\mbox{perimeter}\left(\triangle v_{i}v_{j}v_{k}\right)}{\mbox{area}\left(\triangle v_{i}v_{j}v_{k}\right)} $$no less than $$\min_{v_{0},v_{1},v_{2},v_{3},v_{4}\in\partial\mathbf{D}}\max_{0\le i,j,k\le4}\frac{\mbox{perimeter}\left(\triangle v_{i}v_{j}v_{k}\right)}{\mbox{area}\left(\triangle v_{i}v_{j}v_{k}\right)}? $$

I think it is not hard to see that the second quantity is minimized by the regular pentagon. First, using Günter's observation, the problem is equivalent to find $$\max_{v_{0},v_{1},v_{2},v_{3},v_{4}\in\partial\mathbf{D}} \min_{0\le i,j,k\le4} \mbox{inradius}(v_i,v_j,v_k).$$ If we don't have a regular pentagon, then there are 3 points who are contained in an arc whose length is strictly less than $4\pi/5$, suppose they are $v_1,v_2,v_3$ in this order. The $\mbox{inradius}(v_1,v_2,v_3)$ is maximized if $v_2$ is halfway between $v_1$ and $v_3$. So this quantity will be less than the inradius of three consecutive vertices of the pentagon, which is less than the inradius of three nonconsecutive vertices of the pentagon. 

