Prove $4n < n^2 - 7$ for $n$ is greater than or equal to $6$

Question

We are supposed to be proving this by induction and I know the basis is true $4(6) < 36-7$ and the inductive hypothesis is $4n<n^2-7$ for n $ \ge $6 but I am not sure what the next step is. Do I substitute $n$ for $k+1$? or add $k+1$ on both sides? Thanks!

Answer 1

Add $-4n+11$ to each side. We need to prove $$(n-2)^2>11$$

It is enough to stop here. But if you are looking for induction, then,

CLAIM

$$(n-2)^2>11$$ Is true for all $n \ge 6$.

Base Case It is true for $6$.

Now, Assume that it is true for some $k \ge 6$. Realize that $(k-2)^2$ is a increasing function for $k \ge 2$.

Thus, we have that $(k+1-2)^2 \ge (k-2)^2>11$

Answer 2

Notice that $(n - 2)^2 > 11$ for $n \geq 6$.

Answer 3

If you've shown the base case, you have

$$4n < n^2 - 7$$

$$4n + 4 < n^2 +2n - 6$$

$$4n < n^2 - 7 + 2n - 3$$

Since we already know $4n < n^2 - 7$, we only need to know $2n - 3 \ge 0$, which is true for $n \ge 2$

Answer 4

I think that, as others have shown, $(n- 1)^2> 11$ for $n\ge 6$ is better than "induction" but if you are required to do that, if this is true for b= k, that is, if 4n< n^2- 7, then (n+ 1)^2- 7= n^2+ 2n+ 1- 7> 4n+ 2n+ 1. Then, with $n\ge 6$, $2n\ge 12$ and $2n+ 1> 13$. So $(n+1)^2- 7> 4n+ 2n+ 1> 4n+ 13n> 4n+ 4> 4(n+ 1)$

Answer 5

Let $P(n)$ be the statement that $4n < n^2 - 7$.

You have shown that $P(6)$ holds. Assume $P(m)$ holds for some integer $m \geq 6$. Then $\color{blue}{4m < m^2 - 7}$. Let $n = m + 1$. Then \begin{align*} 4(m + 1) & = \color{blue}{4m} + 4\\ & < \color{blue}{m^2 - 7} + 4 && \text{by the induction hypothesis}\\ & < m^2 + 13 - 7\\ & \leq m^2 + 2m + 1 - 7 && \text{since $m \geq 6 \implies 2m + 1 \geq 13$}\\ & = (m + 1)^2 - 7 \end{align*} Since $P(6)$ holds and $P(m) \implies P(m + 1)$ whenever $m \geq 6$, $P(n)$ holds for each integer $n \geq 6$.