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