I am trying to find the lower bound of the ratio between the sum of diagonals of some convex pentagon $S_d=d_1+d_2+d_3+d_4+d_5$ and its perimeter $P=l_1+l_2+l_3+l_4+l_5$.
After some experimentation by trial-and-error method, the minimum ratio I have achieved is close but above of $\frac{S_d}{P}\approx \frac{3}{2}$, so I conjecture that $$\frac{S_d}{P}>\frac{3}{2}$$
However, I am not able to prove if this ratio is correct, or can be improved further. Applying the Law of Cosines, I get that $$\frac{\sum_{i=1}^{5}{d_i}^2}{2}=\sum_{i=1}^{5}\left({l_i}^2-\prod{l_il_{i+1}}\cos\theta_i\right)$$
Where $\theta_i$ is the interior angle between $l_i$ and $l_{i+1}$. It is also known that $\sum_{i=1}^{5}{\theta_i}=540^º$. As the cosine of any angle between $90^°$ and $180^°$ is negative, and it is easy to prove that at least two interior angles of every convex pentagon must be in this range, I guess that the second term in the RHS (between brackets) of the equality might be positive but relatively close to zero; however, I have no clue about how to bound it properly.
As it is easy to show that $P<S_d$, we can derive from this that $$\frac{S_d}{P}>1$$
I am not able to derive anything more useful from the above equality. Any help on this would be welcomed.
Thanks!
$\frac{S_d}{P}>\frac{3}{2}$ is false.
Indeed, let $ABCDE$ be a convex pentagon. If $A,B,C,D$ are fixed and $E$ moves "a lot", that ratio must decrease.
The limiting case would be a triangle $ACD$ with points $B\in AC$ and $E\in DA$.
Inspecting with Geogebra, that quotient can be as close to $1$ as you like.
On the other hand, what does make sense to me is to find an upper bound, which I suspect will be reached in the regular pentagon (it is known to be $\Phi$, the golden number)