This question follows from another question I've asked on triangles Length of sides of a triangle and area
Let $P_1$ and $P_2$ be two convex quadrilaterals such that $P_1\neq P_2$ and $Area(P_1)\ge Area(P_2)$. Is it true that it is not possible that all sides and diagonals of $P_1$ are shorter than the corresponding sides and diagonals of $P_2$?
This sentence seems true to me, I'll explain my reasoning. Suppose that all four sides and one diagonal of $P_1$ are shorter than the corresponding sides and diagonal of $P_2$. Then, as it is explained in the answer to my older question, the two triangles which triangulate $P_2$ should be "flatter" (otherwise it's not possible $Area(P_1)\ge Area(P_2)$). From this it should follow that the other diagonal of $P_1$ is longer than the corresponding diagonal of $P_2$.
Do you think it's correct? Can you help me formalizing it?
Edit: following user Raffaele's hint I checked out the formulas in this wiki page https://en.wikipedia.org/wiki/Quadrilateral#cite_note-10, but none of them seems useful to solve this problem