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. 
 A: Hint:  rewrite the given inequality as:
$$2 n - 18 \lt n^2- 8n + 8 \quad\iff\quad 0 \lt n^2 - 10 n + 26 \quad\iff\quad 0 \lt (n-5)^2+1$$
You can prove the latter by induction if it really needs be, but of course, it's painfully obvious what the direct proof is.
A: 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. 
A: 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 $n^2-10n+26=(n-5)^2+1>0$.
A: When we group the terms on both sides, the expression will become clearer.
Therefore:
$$2(k+1) - 18 < (k+1)^2 - 8(k+1) + 8$$
$$(2k + 2) - 18 < (k^2 + 2k + 1) + (-8k-8) + 8$$
$$(2k) + (2-18) < (k^2) + (2k-8k) + (1-8+8)$$
$$2k - 16 < k^2 - 6k + 15$$
$$0 < k^2 - 8k + 31.$$
Now convert $k^2 - 8k + 31$ into vertex form and show that the roots are not integers, which means that no integers satisfy the condition. Additionally, what is the discriminant the of the quadratic? What does this tell you?
