inequality - Prove that $frac{1}{2}cdot frac{3}{4} cdots frac{2n-1}{2n} leq frac{1}{sqrt{3n+1}}$ without induction

Prove that $\frac{1}{2}\cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n+1}}$

I know this can be easily proved by induction. But I am looking for another approach. How do I prove this without induction?
Here this question exists - How does one prove that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}?$. But the only one solution there uses induction. But I am looking for solution other than induction.

Answer

If we consider
$$a_n = \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n}=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}$$
we have:
$$a_n^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right) = \frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{2}$$
hence:
$$4n a_n^2 \leq \exp\sum_{k=2}^{n}\frac{1}{4k(k-1)} \leq \exp\left(\frac{1}{4}\right) \tag{3}$$

and:
$$a_n \leq \sqrt{\frac{1}{4e^{-1/4}n}} \tag{4}$$
is a stronger inequality, since $4e^{-1/4}\approx 3+\frac{1}{9}$.
No induction, just squaring and creative telescoping.