Induction on natural numbers My textbook, Logic and Discrete Mathematics by Grassman and Tremblay, has an example which I can't wrap my head around (example 3.4; page 127). It shows that for all $n$, $2(n+2)\le(n+2)^2$. As the inductive base, we have $P(0): 2(0 + 2) \le (0 + 2)^2 = 4 \le 4$, which is true. We assume $P(n): 2(n+2)\le(n+2)^2$ is true as the inductive hypothesis. But for the inductive step, the author does: $$\begin{align}2((n + 1) + 2) &= 2((n + 2) + 1) \\\ &= 2(n + 2) + 2 \\\ &\le (n + 2)^2 + 2 \\\ & \le n^2 + 4n + 6 \\\ & \le n^2 + 6n + 9 \tag 5\\\ & = ((n+1)+2)^2\end{align}$$ The first four lines are straightforward enough. But on line 5, the $2n + 3$ is added and the author doesn't state where that came from. Can someone please explain what happened in that step?

Because it's not specified that we're working with natural numbers, you should probably take induction on all integer. If it's so then you should take in account only the second part of the proof.

Note that the LHS is negative for $n < -2$ and the RHS is always positive as a square. So you the basis should start from $n = -2$ instead.

Note that for numbers $n \ge -\frac 32$ the following statment is always true:

$$2n + 3 \ge 0$$

Our inductive step starts from $n=-1$ so it this inequality holds for every number.

Now just add $n^2 + 4n + 6$ to both sides. It won't change nothing, beacuse we are adding it to both sides.

$$n^2 + 4n + 6 + 2n + 3 \ge n^2 + 4n + 6$$ $$n^2 + 6n + 9 \ge n^2 + 4n + 6$$