Show that functions in O(1) can't grow faster than their composition with themselves Let $f(n)$ be a function s.t $f(n)\geq 1 $ for every $n$.

I want to disprove that if  $f(n) = \omega (f(f(n)))$
 then it means that  $f(n) = O(1)$.

I thougt of 2 approaches to show that this statement is incorrect:


Give an Example of a function that $f(n) = \omega (f(f(n)))$ for it,


for example: $f(n) = \sqrt n$

and then I'm supposed to get a contradiction, and show that $\sqrt n \neq O(1)$
but that's the step I'm stuck at, but I'm not sure how that contradiction is derived. I mean formally, because I understand it by intuition.


Start with the function that suffice $f(n) = O(1)$  and then show that it's impossible to get the condition under that function (It's sort of going from the end to the beginning)


what's the right approach and how do I get it right?
To complete the first approach and show that $\sqrt{n}\notin O(1)$, you just need to show that the statement "There exist constants $c$ and&nbsp;$n_0$ such that, for all $n\geq n_0$, $\sqrt{n}\leq c\cdot1$" is false. That is, you need to show that, for any $c$ and&nbsp;$n_0$, there is an $n\geq n_0$ such that $\sqrt{n}&gt;c$.  For example, you can take $n=n_0+c^2$ (or $\max\,\{n_0,c^2\}$, if you prefer).

I don't really understand your second approach. If $f(n)=O(1)$, it's perfectly possible that $f(n)\in \omega(f(f(n)))$: take any constant function, for example.