The Collatz's function is a mapping $f:\mathbb{Z}_+\to\mathbb{Z}_+$ satisfying \[ f(x)=\begin{cases} 3x+1,& \mbox{as }x\mbox{ is odd}\\ x/2, & \mbox{as }x\mbox{ is even.}\\ \end{cases} \] In addition, let us define the notation $f^1=f$ and inductively $f^{k+1}=f\circ f^k,$ or to say in another words, $f^k(x)=\underbrace{f(\ldots (f}_{k\text{ times}}(x)\ldots ).$ Prove that there is an $x\in\mathbb{Z}_+$ satisfying \[f^{40}(x)> 2012x.\]
Problem
Source: Finland 2012, Problem 5
Tags: function, induction, number theory unsolved, number theory