Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Özgür Cem Birler)
Prove that $\prod\limits_{i=1}^n \frac{2i-1}{2i} \leq \frac{1}{\sqrt{3n+1}}$ for all $n \in \Bbb Z_+$ 
I examined the solutions and tried to apply the methods used there to make sure whether I understand it or not. I'm concerned about the step of induction.
A task from an earlier exam at my university:

Prove by induction:
$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$

Attempt:
rewritten:
$$\prod_{i=1}^n\frac{4i-1}{4i+1}<\sqrt{\frac{3}{4n+3}}$$
$(1)$ base case: $\tau(1)$
$$\frac{3}{5}=\sqrt{\frac{4}{7}}<\sqrt{\frac{3}{7}}$$
$(2)$ assumption: 
Let:$$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$$
hold for some $n\in\mathbb N$
$(3)$ step: $\tau(n+1)$
$$\frac{4n+3}{4n+5}\cdot\prod_{i=1}^n\frac{4i-1}{4i+1}<\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\frac{\sqrt{3(4n+3)}}{4n+5}<\sqrt{\frac{3}{4n+7}}$$
$$\frac{12n+9}{16n^2+40n+25}<\frac{3}{4n+7}\iff\frac{48n^2+120n+63-48n^2-120n-75}{\underbrace{(4n+5)^2(4n+7)}_{>0}}<0\iff-\frac{12}{(4n+5)^2(4n+7)}<0$$
Is this combined legitimate?
 A: Your calculations are correct. 
But I thought it might be helpful also to mention another nice trick to handle such products:


*

*Set $A = \frac 35 \cdot \frac 79 \cdots \frac{4n-1}{4n+1}$

*Let $B = \frac 57 \cdot \frac 9{11} \cdots \frac{4n+1}{4n+3}$
It follows immediately
$$A < B \Rightarrow A^2 < AB = \frac 3{4n+3}$$
Done. 
A: I concur with everyone else that it's basically right. Everyone has their own style, but the following is how I would probably write up the "meat and potatoes" of the proof:

Let's start with some preliminary observations. Note that
\begin{align}
\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}
&=\frac{4n+3}{4n+5}\cdot\frac{\sqrt{3}}
{\sqrt{4n+3}}\\[1em]
&=\frac{\sqrt{4n+3}\cdot\sqrt{3}}{4n+5}\\[1em]
&= \sqrt{\frac{(4n+3)(3)}{(4n+5)^2}}\\[1em]
&= \sqrt{\frac{12n+9}{16n^2+40n+25}}.
\end{align}
More concisely, we have
$$
\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}=\sqrt{\frac{12n+9}{16n^2+40n+25}}.\tag{1}
$$
As another observation, note that
\begin{align}
\frac{12x+9}{16x^2+40x+25} < \frac{3}{4x+7}
&\iff \frac{12x+9}{16x^2+40x+25} - \frac{3}{4x+7} < 0\\[1em]
&\iff \frac{(48x^2+120x+63)-(48x^2+120x+75)}{(4x+5)^2)(4x+7)}<0\\[1em]
&\iff \frac{-12}{(4x+5)^2(4x+7)}<0\\[1em]
&\iff \frac{12}{(4x+5)^2(4x+7)}>0\\[1em]
&\iff x\in\Bigl(-\frac{7}{4},-\frac{5}{4}\Bigr)\cup\Bigl(-\frac{5}{4},\infty\Bigr).
\end{align}
More specifically, note that, for any natural number $n$, it follows from above that
$$
\frac{12n+9}{16n^2+40n+25} < \frac{3}{4n+7}.\tag{2}
$$

Now your proof can flow very naturally:
\begin{align}
\prod_{i=1}^{k+1}\frac{4i-1}{4i+1}
&=\prod_{i=1}^k\frac{4i-1}{4i+1}\cdot\frac{4(k+1)-1}{4(k+1)+1}\\[1em]
&<\frac{4k+3}{4k+5}\cdot\sqrt{\frac{3}{4k+3}}\\[1em]
&=\sqrt{\frac{12k+9}{16k^2+40k+25}} & \text{(by $(1)$)}\\[1em]
&<\sqrt{\frac{3}{4k+7}} & (\text{$\sqrt{x}$ strictly increases and by $(2)$})\\[1em]
&= \sqrt{\frac{3}{4(k+1)+3}}. & \text{(desired conclusion)}
\end{align}
Maybe that's overdrawn and slightly verbose, but that's how I would write it up if I were doing it for a class.
A: Base case: 
$n = 1$
$\frac {3}{5} < \sqrt {\frac {3}{7}}$
Since both are positive, we can square both sides.
$\frac {9}{25} < \frac {3}{7}$
Cross multiply
$63 < 75$
Inductive hypothesis
Suppose, $\prod_\limits{i=1}^{n} \frac {4i-1}{4i+1} < \sqrt {\frac {3}{4n+3}}$
We must show that when the hypothesis holds, $\prod_\limits{i=1}^{n+1} \frac {4i-1}{4i+1} < \sqrt {\frac {3}{4n+7}}$
$\prod_\limits{i=1}^{n+1} \frac {4i-1}{4i+1}  = \frac {4n+3}{4n+5}\prod_\limits{i=1}^{n} \frac {4i-1}{4i+1} < \frac {4n+3}{4n+5}\sqrt {\frac {3}{4n+3}}$ based on the inductive hypothesis. 
$\frac {4n+3}{4n+5}\sqrt {\frac {3}{4n+3}} = \sqrt {\frac {3(4n+3)}{(4n+5)^2}}$
$(4n+5)^2 = (4n+3)(4n+7) + 4$
$\sqrt {\frac {3(4n+3)}{(4n+5)^2}} = \sqrt {\frac {3(4n+3)}{(4n+3)(4n+7) + 4}} < \sqrt {\frac {3(4n+3)}{(4n+3)(4n+7)}} = \sqrt {\frac {3}{4n+7}}$
Which is what we were required to show.
A: It is correct.
Anyways you already ended the proof when you stated that
$\frac{4n+3}{4n+5}\prod_{i=1}^n\frac{4i-1}{4i+1}$ $<$ $\sqrt{\frac{3}{4n+7}}$
