Prove $n - 2 < \frac{n^2 - n}{12}$ by Mathematical Induction

Question

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.

Answer 1

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

Answer 2

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.