Let $X_1, ..., X_5$ be the vertices of a convex pentagon with perimeter $p$ with its centroid at the origin, satisfying $d(X_i, X_j) < \frac{p}{3}$ for $1 \leq i,j \leq 5$, where $d$ is the Euclidean distance. Let $X_m$ be a vertex of that pentagon with maximal distance to the origin and let $P$ be the orthogonal projection onto the straight line through $X_m$ and the origin.
My question is: Is it true that
$$\sum_{i=1}^5 d(X_m,PX_i) \leq \frac{1}{2\sin\left(\frac{\pi}{5}\right)}\sum_{i=1}^5 d(X_m,X_i)$$
and if yes, how could I prove it?
A positive answer (and a proof) would provide a way to prove the result in this MSE post, because we can establish a lower bound for the left hand side of our inequality involving $\frac{1}{2\sin\left(\frac{\pi}{5}\right)}$. The $\frac{p}{3}$ condition should guarantee that the pentagon is in some sense not too far away from a regular pentagon, so that on average the distance to the projected points is sufficiently smaller than the distance to their non-projected versions. I have verified the inequality for several pentagons (satisfying the $\frac{p}{3}$ condition) and it is holding up so far. Note that it's invariant under scaling, so we could assume something like $p = 1$ if necessary. We could also instead assume wlog that $X_m = (1, 0)$, then if $X_i = (a_i, b_i)$ for all $i$, we get $PX_i = (a_i, 0)$ and the inequality becomes
$$\sum_{i=1}^5 \vert a_i-1 \vert \leq \frac{1}{2\sin\left(\frac{\pi}{5}\right)}\sum_{i=1}^5 \sqrt{(a_i-1)^2 + b_i^2}.$$
I'm looking primarily for a non-computer-assisted proof, but any kind of help is appreciated.