<s>[INST] How does one prove that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}?$ I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:
$\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$
I would appreciate your help.
 [/INST] You want to show that
$$
\frac1{\sqrt{3n+1}}\;\frac{2n+1}{2n+2}\leq\;\frac1{\sqrt{3n+4}}.
$$
Since everything is positive, this inequality is the same as
$$
(3n+4)(2n+1)^2\leq (3n+1)(2n+2)^2.
$$
After expanding and cancelling the $n^3$ terms we get
$$
12n^2+19n\leq 24n^2+20n.
$$
This inequality holds trivially for any $n\in\mathbb{N}$, and now you can retrace the steps back to your desired inequality. 
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