I'd like to prove the statement:
$\forall c,m\in \mathbb{R}^+,\exists n \in \mathbb{N},n≥m ∧ n^2-2n-1>c(n+3)$
I was thinking of rearranging the equation to isolate for $n$ but it gets really messy, is that really the right approach?
Otherwise, how should I approach determining the value of $n$ for this statement?
Set $x$ as the point where the quadratic $f(x) = x^2 - 2x - 1$ crosses the line $g(x) = c(x+3)$ where $x>0$. We know that $x$ exists as the inflection point of $f$ is at $(1,-2)$ which is below $g$ for any $c>0$.
Then put $n = \max(m, \lceil x \rceil + 1)$ where $\lceil \cdot \rceil$ is "round up to next integer".
Then by construction, $n \geq m$ and $n^2 - 2n - 1 > c(n+3)$, as required.
The key idea is that a quadratic will eventually (asymptotically) exceed any line, so one can choose $n$ to satisfy the constraint from $c$. Then use $\max$ to satisfy the constraint from $m$.