<s>[INST] 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?
 [/INST] 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.
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