Prove $n - 2 < \frac{n^2 - n}{12}$ by Mathematical Induction I am trying to prove the following $n - 2 < (n^2 - n)/12$ when $n > 10$ by Mathematical Induction. The following is what I've come up so far (please bare with me):
Property to be proven $P(n)$:
$$
n - 2 < (n^2 - n)/12 \hspace{.5cm}\leftarrow P(n)
$$
[For now I am assuming to solve for integer values, thus for the basis step I've used] Show that $P(11)$ is true:
$$
11 - 2 < (11^2 - 11)/12 \hspace{.5cm} \leftarrow \text{basis } P(11)\\
9 < 110/12 \\
108/12 < 110/12
$$
Hence $P(11)$ is true.
Show that for every integer $k \geq 11$, if $P(k)$ is true then $P(k + 1)$ is also true:
Suppose that $k$ is any integer with $k \geq 11$ such that
$$
k - 2 < (k^2 - k)/12. \hspace{.5cm} \leftarrow P(k) \text{ inductive hypothesis}
$$
[We must show that $P(k + 1)$ is true. That is:] We must show that
$$
(k + 1) - 2 < ((k + 1)^2 - (k + 1))/12. \hspace{.5cm} \leftarrow P(k + 1)
$$
or, equivalently,
$$
k - 1 < (k^2 + k)/12.
$$
or, also,
$$
12k - 12 < k^2 + k
$$
Now, from the inductive hypothesis:
$$
k - 2 < (k^2 - k)/12 \\
12(k - 2) < k^2 - k \hspace{.5cm} \text{multiply the inequality by 12} \\
12k - 24 < k^2 - k \\
(12k - 24) + 2k < (k^2 - k) + 2k \hspace{.5cm} \text{add } 2k \text{ on both sides}\\
(12k - 12) + (2k - 12) < k^2 + k \hspace{.5cm} \text{reordering and grouping}
$$
Because $2k - 12 > 0$ since $k \geq 11$.
$$
k^2 + k > 12k -12
$$
[as was to be shown.]
At this time, I am unsure whether the statement "Because $2k - 12 > 0$ since $k \geq 11$." allows me to complete the proof. Also, I'm unsure how to proceed otherwise.
I hope to obtain feedbacks from everyone in regards of this proofing.
Thank you in advance, and have a nice day.
 A: How about this
For $n=11$, it is correct
If for a given $n$ the inequality is correct, we are going to prove that the inequality is correct for $n+1$
$\frac{(n+1)^{2}-(n+1)}{12}=\frac{n^{2}-n}{12}+\frac{2n}{12}$
For $n>10$, $\frac{2n}{12}>1$
Therefore,
$\frac{(n+1)^{2}-(n+1)}{12}>\frac{n^{2}-n}{12}+1$
$\frac{(n+1)^{2}-(n+1)}{12}>n-2+1$
$\frac{(n+1)^{2}-(n+1)}{12}>(n+1)-2$
Thus proving the inequality is correct for all integers $n>10$ by induction
A: Another way to prove
an inequality of the form
$f(n) < g(n)$
for $n \ge n_0$
is to
(1) show that $f(n_0) < g(n_0)$
and
(2) if $n \ge n_0$ then
$f(n+1)-f(n) \le g(n+1)-g(n)
$.
This says that $g(n)$ increases
at least as fast as $f(n)$
so $f(n)$ can never catch up.
In this case
$f(n+1)-f(n) = 1$,
so we need to show that
$g(n+1)-g(n) \ge 1$.
Here,
$g(n+1)-g(n)
=\dfrac{(n+1)^2-(n+1)}{12}-\dfrac{n^2-n}{12}
=\dfrac{(n+1)^2-n^2-((n+1)-n)}{12}
=\dfrac{2n}{12}
=\dfrac{n}{6}
$
so
$f(n+1)-f(n)
\le g(n+1)-g(n)
$
for $n \ge 6$.
Since
$f(11) < g(11)$,
$f(n) < g(n)$
for $n \ge 11$.
The advantage of this method
is that we do not need to consider
$g(n)-f(n)$,
so the calculations are often simpler.
