Q: Minimal area of quadrilateral divided by its shorter diagonal Given four sides $a,b,c,d$, which konvex quadrilateral with these sides minimizes $\frac{\sqrt{A}}{f}$, where $A$ denotes the area and $f$ the shorter diagonal.
My guess is, that it is minimized by the cyclic quadrilateral, due to the case where all sides have the same length but I have no clue if this is true for the general case and how to prove it.

A: The problem has no simple answer. To see this consider two easy solvable examples:

*

*Orthodiagonal quadrilateral: $a^2+c^2=b^2+d^2$. In this case the minimizing quadrilateral is that one with equal diagonals (or with the diagonal ratio possibly closest to $1$ if the ratio $1$ is not allowed by the convexity requirement).


*"Degenerable" quadrilateral: $a+b=c+d$, $a\ne d$. In this case the minimizing quadrilateral is the degenerate one with the minimizing value being $0$.
Generally the minimizing quantity is
$$
\frac{2A}{q^2}=\frac pq\sin\theta,
$$
where $p$ and $q$ are the longer and the shorter diagonals, respectively, and $\theta$ is the angle between the diagonals. Obviously for a constant angle (as in the first example) the minimum will be achieved at $p=q$, but in some cases (as in the second example) the decrease of the angle $\theta$ can be decisive.
