Let $n\in\mathbb{N}\setminus\{1,2\}.$ Let $a_1,a_2,\ldots,a_n$ be $n$ (not necessarily distinct) real numbers each in the interval $(1,F_n),$ where $F_n$ is the $n^{\text{th}}$ Fibonacci number. Show that we can pick three numbers $a_i,a_j,a_k$ from the $n$ real numbers such that there exists a triangle with lengths $a_i,a_j$ and $a_k.$
I initially thought of using some sort of pigeonhole principle, but that doesn't seem to work. I have no idea how to proceed. Of course, the triangle inequality is key here, but like I said, I don't know how to use it.
For $n=3$ it is trivial. We have $a_1,a_2,a_3\in(1,2).$ WLOG, assume $1<a_1\leq a_2\leq a_3<2.$ It is obvious that $a_2+a_3>a_1.$ It is also obvious that $a_3+a_1>a_2.$ We need to show that $a_1+a_2>a_3.$ Suppose not. Then, $a_1+a_2\leq a_3.$ But then, the LHS is greater than $2$ while the RHS is less than $2.$ A contradiction.
Maybe this approach can be generalised? I don't know.
You have proved the problem for $n=3$.
For $n=4,$ consider the sequence $1<a_1\leq a_2\leq a_3\leq a_4 <3=F_4$. If $a_3 <2=F_3,$ then by the case $n=3$, $a_1,a_2,$ and $a_3$ form a triangle. So, let's assume $a_3 \geq 2.$ In this way, we have:
$$a_2+a_3>1+2=3>a_4,$$
implying that $a_2,a_3,$ and $a_4$ form a triangle.
Now, let's assume that we can pick three numbers (forming a triangle) out of every sequence of length $n$ or $n+1$ that satisfies the requirement of the problem.
Consider the sequence $1<a_1\leq a_2\leq . . . \leq a_n \leq a_{n+1} \leq a_{n+2} <F_{n+2}.$
If $a_n <F_n$, by assumption there are three real numbers among $a_1, a_2, ... ,a_n$ that can form a triangle. The same holds if $a_{n+1}<F_{n+1}.$ Hence, we may assume that $a_n \geq F_n$ and $a_{n+1} \geq F_{n+1}$. However, in this way, we have:
$$a_n+a_{n+1} \geq F_n+F_{n+1}=F_{n+2}>a_{n+2},$$
implying that $a_n,a_{n+1},$ and $a_{n+2}$ form a triangle.