find a pentagon where you cannot get a triangle from any diagonals (a pentagon may not be convex) (a triangle is not in a pentagon; it is obtained with the help of diagonals; if there are any three diagonals of length which make it possible to form a triangle from these three diagonals, then that pentagon already considered unnecessary to us) I don't know how to find that pentagon. I just know that if it's convex, then there always is such a triangle of diagonals. So I have to look for a not convex pentagon.
When you also allow outer diagonals, then the diagonals form a loop. By deforming the loop you also get a pentagon. So the question becomes can we find a pentagon such that the lengths of the sides never form a triangle, and this pentagon are the diagonals from a different pentagon. We can find such lengths: $1$, $2$, $3$, $5$, $8$ (note that this is part of the Fibonacci sequence) Below is such a pentagon (red are the diagonals) The lengths of $AB$, $BC$, $CD$, $DE$, and $EA$ are respectively $8$, $5$, $3$, $2$, and $1$.