How do I show this, possibly using the pigeonhole principle?

Question

Show that if you choose any $12$ real numbers between $1$ and $12$, three of them must be the sides of an acute triangle.

Answer 1

Note that three numbers $a\leq b\leq c$ are the sides of an acute triangle iff $a^2+b^2>c^2$. Suppose no triple of $d_i$ among $1< d_1,\leq \cdots\leq d_{12}< 12$ are the sides of an acute triangle. Then $1<d_1^2\leq \cdots \leq d_{12}^2<144$, and we have $1<d_1^2,1<d_2^2$ and $d_{i}^2+d_{i+1}^2\leq d_{i+2}^2$. But these last three inequalities imply that $d_i^2>F_i$, the $i^{th}$ Fibonacci number, and $F_{12}=144$. Thus we have a contradiction.