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!

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.
