What is the maximum fraction of the product of the diagonals of a quadrilateral compared to the product of its sides?
Let the sides be denoted by the lowercase letters $a, b, c, d, e$, and the non-adjacent diagonals by the corresponding uppercase letters $A, B, C, D, E$. This way, we have counted each diagonal and each side exactly once. For each possible cyclic quadrilateral determined by the five vertices, we write down Ptolemy's theorem, which gives us the following equations:

$$
b e + a A = C D, \quad c a + b B = D E, \quad d b + c C = E A, \quad e c + d D = A B, \quad a d + e E = B C
$$

The product of these equalities, after some rearrangement, takes the form

$$
\left(1+\frac{a A}{b e}\right)\left(1+\frac{b B}{c a}\right)\left(1+\frac{c C}{d b}\right)\left(1+\frac{d D}{e c}\right)\left(1+\frac{e E}{a d}\right)=\left(\frac{A B C D E}{a b c d e}\right)^{2}
$$

On the right-hand side, the ratio we are asked for in the problem appears. Our goal now is to estimate the left-hand side with the same ratio, hopefully leading us to the desired extremum.

Instead of the product on the left-hand side, consider more generally the expression

$$
\left(1+\lambda_{1}\right)\left(1+\lambda_{2}\right) \ldots\left(1+\lambda_{n}\right)
$$

with positive $\lambda_{i}$ (in our case, $n=5$), which, when expanded, is a polynomial of degree $n$ in the $\lambda_{i}$. Apply the inequality between the arithmetic and geometric means to the sum of the terms of degree $m$ for every $1 \leq m \leq n$: we get that the corresponding sum does not increase if we replace each $\lambda_{i}$ with $\sqrt[n]{\lambda_{1} \lambda_{2} \ldots \lambda_{n}}$ (the reason for this is that the individual sums - the homogeneous components of the polynomial we start with - are symmetric in the $\lambda_{i}$). Therefore, the product itself does not increase if we replace each $\lambda_{i}$ in this way, i.e.,

$$
\left(1+\lambda_{1}\right)\left(1+\lambda_{2}\right) \cdot\left(1+\lambda_{n}\right) \geq\left(1+\sqrt[n]{\lambda_{1} \lambda_{2} \cdot \lambda_{n}}\right)^{n}
$$

With the notation $h=\sqrt[5]{\frac{A B C D E}{a b c d e}}$, the original (1) equality together with (2) gives the inequality

$$
h^{10} \geq(1+h)^{5}
$$

Since the function $x \mapsto x^{5}$ is strictly increasing on the real numbers and $h$ is positive, the above inequality means that $h \geq \frac{\sqrt{5}+1}{2}$, i.e.,

$$
\frac{a b c d e}{A B C D E}=h^{-5} \leq\left(\frac{\sqrt{5}+1}{2}\right)^{-5}=\left(\frac{\sqrt{5}-1}{2}\right)^{5}=\frac{5 \sqrt{5}-11}{2}
$$

It is clear that the above inequalities hold with equality in the case of a regular pentagon, so the maximum of the ratio $\frac{a b c d e}{A B C D E}$ is $\frac{5 \sqrt{5}-11}{2}$.

Based on the work of Ákos Gröller (Fazekas M. Főv. Gyak. Gimn., IV. o.t.)

Remark. The inequality (2) is equivalent to the fact that the function $f(x)=\ln \left(1+e^{x}\right)$ is convex, which follows, for example, from the fact that its derivative $f^{\prime}(x)=1-\frac{1}{1+e^{x}}$ is increasing.