If $a_i$ are not necessarily distinct real & $1<a_i<13$ for all $i$.Show that we can choose three of them such that they are lengths of a triangle

Question

If $a_1,a_2,...a_7$ are not necessarily distinct real numbers such that $1<a_i<13$ for all $i$ .

Show that we can choose three of them such that they are lengths of a triangle

I tried to solve this problem in the following steps:

Using the principle:

Sum of length of two sides > Length of third side

1. Sort all the $a_i$s in non descending order.

2. Observe that if $a_i,a_{i+1},a_{i+2}$ do not form a non-degenerate triangle,then $a_i+a_{i+1}\le a_{i+2} \implies$ as $ a_i$s are all sorted,then $a_{i-1}\le a_i \implies a_{i-1}+a_{i+1}\le a_{i+2}\implies$ $a_{i-1},a_{i+1}, a_{i+2}$ do not form a non-degenerate triangle.Again $a_i+a_{i+1}\le a_{i+2}\implies a_i+a_{i+1}\le a_{i+3}$. Thus $a_i,a_{i+1}, a_{i+3}$ do not form a non-degenerate triangle.

Now what should be my next step?Please help.

Answer 1

Well, if the numbers were $1,1,2,3,5,8,13,$ you couldn't form a triangle. I think you should try to prove that if $k_n$ is the smallest real number such that $1<a_i<k_n, 1\le i\le n$ guarantees that it is possible to choose $3$ of the $a_i$ that form a triangle, then $k_n$ is the $n$th Fibonacci number.

EDIT

While you were kind enough to say this is a brilliant observation, it isn't so. If it were, I couldn't have made it. I just made small examples. What is the bound for $3$ numbers? For $4$ numbers? Then I saw the pattern.