Question We consider the set $A_n=\{a_1,a_2,...,a_{2n}\}$ where its elements are different real numbers and strictly positive. Show that for any natural number $n\geq2$ we can choose $4$ different numbers $a,b,c,d$ from $A_n$ so that $| ad-bc | < \frac{bd}{n-1}$.
My idea: The first thing that came into my mind when I saw the inequality we have to prove was the Lagrange equality or the means inequality.
I tried using them but I didn't get to any right answer.
I also tried to square to get rid of the absolute value function but it didn't help me in any way.
Hope one of you can help me! Thank you!
I will try to give more details as requested. First of all, we may renumber the indexes of the elements of $A$ such that $a_1 < a_2 < \ldots < a_{2n}.$ Because there are $2n$ numbers, there are $\frac{2n}{2} = n$ quotients $$\frac{a_1}{a_2}, \frac{a_3}{a_4}, \ldots, \frac{a_{2n-1}}{a_{2n}}. \tag{a}\label{a}$$ Because $0 < a_1 < a_2$, we have $0 < \frac{a_1}{a_2} < 1$ and similar for the rest of the quotients. Hence, these quotients are in the interval $(0,1)$. In a similar way, we may label these $n$ quotients and then we have $$0 < b_1 < b_2 < \cdots < b_{n} < 1$$ where each of the $b_i$ corresponds to exactly one quotient in the list \eqref{a}.
The argument now goes that there is at least one pair $b_j, b_{j+1}$ such that the distance between these quotients is $$b_{j+1} - b_j < \frac{1}{n+1}.$$ Indeed, if this was not the case, and all the pairs $b_n,b_{n-1}$ ; ... ; $b_2,b_1$ are at a distance of $b_{i+1} - b_i \geq \frac{1}{n+1}$, then the sum of all these distances is $$b_n - b_1 = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2}) + \ldots + (b_3 - b_{2}) + (b_2 - b_1) \geq (n-1)\cdot \frac{1}{n-1} = 1$$ so, $b_{n} - b_1 \geq 1$, which cannot be because $0 < b_1 < b_n < 1$.
It follows that we can find two distinct quotients $b_{j+1} = \frac{a}{b}$ and $b_j = \frac{c}{d}$, with $a,b,c,d \in A$, among \eqref{a} such that $$\left| \frac{a}{b} - \frac{c}{d} \right| < \frac{1}{n+1} \implies \left|ad - bc\right| < \frac{bd}{n+1}.$$