Suppose $a_1,...,a_6$ is a sequence of positive numbers such that $$\max \{a_1,...,a_6\} \le 6 \min \{a_1,...,a_6\}$$ Show that there is some $i,j,k$ with $1\le i < j < k \le 6$ such that $a_i, a_j$ $a_k$ are the sides of a triangle.
This was a problem in a math olympiad Colombia, I couldn't do it and then realised that there was a harder version of it in $2012$ USAMO (and it´s rare since they show this is not a possible case).
I just wanted to show that there exists some $i,j,k$ with $1\le i < j < k \le 6$ such that $a_i + a_j > a_k$, and I did many things, but none of them worked.
Suppose we have a sequence $a_1,...,a_6$ of positive numbers such that
Our goal is to derive a contradiction.
Without loss of generality, assume $$a_1\le a_2 \le a_3\le a_4\le a_5\le a_6$$
By hypothesis, we have $a_6 \le 6a_1$.
Then we get \begin{align*} a_2&\ge a_1\\[4pt] a_3&\ge a_2 + a_1\ge a_1 + a_1=2a_1\\[4pt] a_4&\ge a_3 + a_2\ge 2a_1+a_1=3a_1\\[4pt] a_5&\ge a_4+a_3\ge 3a_1+2a_1=5a_1\\[4pt] a_6&\ge a_5 + a_4\ge 5a_1+3a_1=8a_1\\[4pt] \end{align*} so we have $6a_1\!\ge a_6\ge 8a_1,\;$contradiction.