Prove that $1\cdot f(1)+ 2\cdot f(2)+ ...+ n\cdot f(n) \leq n(n+1)(2n+1)/6$ where $f(n+1) = f(f(n))+1$ Consider checking function $\mathbb{N}\to \mathbb{N}$ relationship $f(n+1) = f(f(n))+1$, for any positive integer $n$.
 Prove that $1\cdot f(1)+ 2\cdot f(2)+ ...+ n\cdot f(n) \leq n(n+1)(2n+1)/6$ for any positive integer $n$.

i am going to try. the defining relation for $f$ gives us the inequality $$nf(n) = n\{f(f(n-1)) + 1\} =n + nf(f(n-1)) \ge n $$ because the range of $f$ consists of positive integers. 
we can use $jf(j) \ge j$ again and again to get
 $$ nf(n) \ge n + nf(f(n-1)) \ge n + nf(n-1) \ge n + n(n-1) = n^2$$
therefore
$$1f(1) + 2f(2) + \cdots + nf(n) \ge 1^2 + 2^2 + \cdots + n^2 = n(n+1)(2n+1)/6 $$
${\bf edit}:$ the argument used to establish $f(n) \ge n$ has a flaw but not the conclusion and it can be fixed.
we already have $f(n) \ge 1 \text{ for all } n \ge 1$ by the range of $f$ being positive integers. now using that and $f(n) = f(f(n-1) + 1$ establishes $f(n) \ge 2 \text{ for } n \ge 2 $. now by induction $f(n) \ge n.$
