Inequality manipulation: $\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}}$ I'm reading Analytic Inequalities by Nicholas D. Kazarinoff. On page 5, we are trying to use induction to prove the inequality
$$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} < \frac{1}{\sqrt{3n + 1}} . $$
For the inductive step, we want to show that it holds for $n + 1$, i.e.
$$ \frac{1}{\sqrt{4n + 5}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 1}{2n} \cdot \frac{2n + 1}{2n + 2} < \frac{1}{\sqrt{3n + 4}} . $$
Kazarinoff says that this is true if
$$ \frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}} $$
is also true. I'm trying to figure out why this is the case, because it's not obvious to me.

Edit: this is completely nonsensical because it's circular.
What I've tried:
\begin{align}
\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} &> \frac{1}{\sqrt{4n + 5}} \\
\implies \frac{1}{\sqrt{4n + 1}} &> \frac{1}{\sqrt{4n + 5}} \cdot \frac{2n + 2}{2n + 1} \\
&< \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} \\
&> \frac{1}{\sqrt{4n + 1}}
\end{align}
That's a pretty useless result that hasn't gotten me anywhere. Any ideas?

By induction assumption, you have
$$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n}.$$
Suppose you have
$$\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}}.$$
Then
$$\frac{1}{\sqrt{4n + 5}}<\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2}<\frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n}\cdot \frac{2n + 1}{2n + 2}.$$
