# 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:

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

Show that $\frac1{\sqrt{3n+1}} \frac{2n+1}{2n+2} \leq \frac1{\sqrt{3(n+1)+1}}$. – user17762 Mar 13 '12 at 18:54
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.