Prove for every natural number $n$, with $n ≥ 7$, that $\frac{(2n−18)}{(n^2−8n+ 8)} < 1$. Using induction, Prove for  every  natural  number $n$, with $n ≥ 7$, that $\frac{(2n−18)}{(n^2−8n+ 8)} < 1$. I cannot get the original equation to match the k+1 equation, and im not sure what I am messing up. I would post my work but I would rather see it done out completely from scratch. Any help appreciated!
 A: Alt. hint (without induction): $\;\dfrac{2n−18}{n^2−8n+ 8} = \dfrac{2(n-7)-4}{(n-1)(n-7)+1} \le \dfrac{2(n-7)-4}{2(n-7)+1}\,$ for $\,n \ge 7\,$.
A: Denominator $n^2-8n+8=n^2-8n+16-8=(n-4)^2-8$. If $n\ge7$ then 
$(n-4)^2-8\ge 3^2-8=1>0$, so then (as suggested in a comment) 
$\frac{(2n-18)}{(n^2-8n+8)}<1\iff (2n-18)<(n^2-8n+8)$. 
That is, prove that $n^2-10n+26>0$ for $n\ge7$, by induction. 
True for $n=7$, we get $49-70+26=5>0$. Assume it is true for some $n$, 
consider $n+1$. $(n+1)^2-10(n+1)+26=n^2+2n+1-10n-10+26=(n^2-10n+26)+2n-9$. Since $2n-9\ge14-9=5>0$, and using the induction hypothesis that $n^2-10n+26>0$ we obtain that $(n^2-10n+26)+2n-9>0$. 
A: Partial Answer as Method of Induction is Not Used.

For curiosity, I have provided an answer that does not use induction.
Multiply the fraction by $1=\dfrac{2n+18}{2n+18}$.
$$\begin{align} \frac{2n-18}{n^2-8n+8}\times\frac{2n+18}{2n+18}&=\frac{4n^2-324}{2n^3 - 16n^2 + 16n + 18n^2 -144n + 144} \\ \\ &=\frac{2n^2-162}{n^3-8n^2+8n+9n^2-72n+72} \\ \\ &= \frac{2n^2-162}{n^3+n^2-64n+72}.\end{align}$$ So now, given that $n\in\mathbb{N}_{\geqslant 7}$, we want to show that $$\begin{align} 2n^2 - 162 &< n^3 +n^2-64n+72 \\ \\ \Leftrightarrow n^2-234&<n^3-64n \\ \\ \Leftrightarrow n^2+64n-234&<n^3.\end{align}$$ So, since $n\geqslant 7$ then there will always exist a constant $c_n$ such that $$n^2 + 64n - (234+n^3-c_n)=0$$ iff (if and only if) $$\begin{align} n&=\frac{-64\pm\sqrt{4096+4(234+n^3-c_n)}}{2} \\ \\ &= 2\left(-16\pm\sqrt{1258+n^3-c_n}\right).\end{align}$$ But since $n>0$ then $$n=2\left(\sqrt{1258+n^3-c_n}-16\right).$$ Of course, to prove the inequality, we need to show that $c_n>0$. Of course $c_n\neq0$ because the constant must always be changing, depending on the value of $n$. Also, given the strict inequality, $c_n\not<0$, thus $c_n>0$.
This proves the inequality.
A: As the denominator $n^2-8n+8$ is positive for $n\ge7,$ the inequality
$$\frac{2n-18}{n^2-8n+8}\lt1$$
is equivalent to the inequality
$$2n-18\lt n^2-8n+8$$
which is equivalent to
$$0\lt n^2-10n+26$$
or
$$0\lt(n-5)^2+1$$
which is true.
(As presented, this is not a proof by induction. However, any proof can easily be converted into a proof by induction, in which the proof of the induction step makes no use of the inductive hypothesis.)
