<s>[INST] 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. 
 [/INST] Notice, $2n-18<n^2-8n+8 \implies n^2-10n+26=(n-5)^2+1>0$.
Let $(n-5)=u$, so that we have $u^2+1>0$. 
Moving over the $1$ gives us $u^2>-1$. We know this is true because all squares of real (and natural) numbers are greater than $-1$.
Therefore, we have $(n-5)^2>-1$, and $(n-5)^2+1>0$, and $n^2-10n+26>0$.
Adding $2n-18$ to both sides gives us $n^2-8n+8>2n-18 \implies 2n-18<n^2-8n+8$.
Thus, our proof is complete. 
</s>