How to prove by induction that for $n$ $ \in \mathbb N $ , $ 2 n - 18 < n^2-8n +8 $? Question:
Prove for $n$ $ \in \mathbb N $ ,  $ 2 n\  -\ 18\ <\ n^2-8n\ +8 $
My attempt:
$ Base\ Case:\ n\ =\ 1,\ it\ holds. $
$I.H:\ Suppose\ 2k-18\ <\ k^2-8k+8,\ where\ k\ is\ a\ natural\ number.$
$ Then,\ \left(k+1\right)^2-8\left(k+1\right)+8\ =\ k^2+2k+1-8k-8+8\ >2k-18+2k+1-8$
I am stuck here. Any help would be appreciated. Thanks. 

Assuming $\quad2n-18 < n^2-8n+8\quad $ then 
$\begin{array}{ll}
2(n+1)-18 = (2n-18)+2 < & (n^2-8n+8)+2\\
&=[n^2-6n+1]-2n+9\\
&=[(n+1)^2-8(n+1)+8]-(2n-9)\\
\end{array}$
So the induction works when $2n-9\ge 0$ that is when $n\ge 5$
You'll get to verify the initial step for $n=5\ :\ \begin{cases}2n-18=-8\\
n^2-8n+8=25-40+8=-7\end{cases}$
For the cases $n\le 4$ you have to verify them manually.
Or go for the algebraic solution $n^2-10n+26=(n-5)^2+1>0$.
