For what values of the natural number $n$ is $2n-8 < n^2-8n+14$? (must use induction)
I have determined that $n$ appears to work for all values except $n=4,5,6$.
I was wondering if this proof that I constructed is valid (for all intents and purposes of this post, I'm skipping the base cases).
Induction Hypothesis Suppose $2n-8 < n^2-8n+14$ is true for all $n\geq 7$.
To show that the proposition is true for the $n+1$th terms, we proceed as follows.
From the induction hypothesis we know
$2n-8 < n^2-8n+14$
Because $2n-6 < 2n-8$, we deduce from the transitive property that
$$2n-6 < n^2-8n+14\tag{1}$$
Because $-8n < -6n$, we have
$-8n < -6n < -6n+7$
$-8n < -6n+7 = -6n+14-7$
$-8n-14 < -6n-7$
Finally, we know that $-6n-7 < -6n+7$. Therefore,
$-8n-14 < -6n+7$
and
$$n^2-8n-14 < n^2-6n+7\tag{2}$$
Using the transitive property on $(1)$ and $(2)$, we conclude
$2n-6 < n^2-6n+7$,
which is the $n+1$th term, as desired.
If you really really want a proof by induction, then I would prove it as follows:
For $n\geq 7$, let $S(n)$ denote the statement $$ S(n) : 2n-8 < n^2-8n+14. $$ Base case ($n=7$): $S(7)$ says that $2(7)-8=6<7=7^2-8(7)+14$, and this is true.
Inductive step: Fix some $k\geq 7$ and assume $S(k)$ to be true where $$ S(k) : 2k-8 < k^2-8k+14. $$ To be shown is that $S(k+1)$ follows where $$ S(k+1) : 2(k+1)-8 < (k+1)^2-8(k+1)+14. $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 2(k+1)-8 &= (2k+2)-8\tag{expand}\\[0.5em] &= (2k-8)+2\tag{rearrange}\\[0.5em] &< (k^2-8k+14)+2\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &< k^2-6k+7\tag{since $k\geq 7$}\\[0.5em] &= (k+1)^2-8(k+1)+14,\tag{desired expression} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 7$. $\blacksquare$