In the following completed proof, in the highlighted steps, it seems as though the proof is defending its claim by showing that $2(\sqrt{k+2}-1) < (\text{the lower highlighted part in the proof})$. Shouldn't it be showing instead that $2(\sqrt{k+2}-1) \ge (\text{the lower highlighted part in the proof})$, in order for the proof to be true? Why or why not?
Why I think this:
Because the P(k+1) we are trying to show, has the left side greater than the right side.
The rest of the work needs to also try to show that the right side(which is being manipulated to try and look like the right side of the P(k+1) inequality) is less than the original P(k+1) inequality's right side, SO THAT the left side of the P(k+1) inequality will always be bigger.
You've agreed that if $a > 1$ and $b>0$, the $ab > b$.
Now lets pick particular $a$ and $b$:
$$a = \sqrt{\frac{4k^2 + 12 k + 9}{4k^2 + 12 k + 8}} \qquad \text{and} \qquad b = 2\sqrt{k+2}.$$
Thus
$$ab = 2 \sqrt{\frac{4k^2 + 12 k + 9}{4k^2 + 12 k + 8}}\sqrt{k+2} > 2\sqrt{k+2} = b.$$
Therefore $ab - 2 > b - 2$ implies
$$2 \left(\sqrt{\frac{4k^2 + 12 k + 9}{4k^2 + 12 k + 8}}\sqrt{k+2}\right) - 2 > 2\sqrt{k+2} - 2 = 2\left(\sqrt{k+2}-1\right).$$