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!
Best Answer
If $a_1 < a_2 < \ldots$, then the quotients $$\frac{a_1}{a_2}, \frac{a_3}{a_4}, \frac{a_5}{a_6}, \ldots$$ are between $0$ and $1$. Pick two quotients whose difference is less than $\frac1{n-1}$ and these are your $a, b, c, d$.