Show that $x_n \leq \frac{1}{\sqrt{3n+1}}$ Let $$x_n=\frac{1}{2} \frac{3}{4}\frac{5}{6}\cdots\frac{2n-1}{2n}$$
Then show that $$x_n \leq \frac{1}{\sqrt{3n+1}}$$ for all $n=1,2,3,\dots$
I try induction but unable to solve this equality.

If  (induction)
$$
x_n \leq \frac{1}{\sqrt{3n +1}}
$$
Then 
$$
x_{n+1} = \frac{1}{2} \frac{3}{4}\frac{5}{6}\cdots\frac{2n-1}{2n}\frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}
$$
You want now to prove that 
$$
\frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{3n+4}}.
$$
That is:
$$
\sqrt{3n+4}(2n+1) \leq (2n+2)(\sqrt{3n+1})
$$
Everything is positive, so the above is true if and only if it is true with the square on both sides. When you so that you get
$$
(3n + 4)(4n^2+4n + 1)\leq (4n^2 + 8n + 4)(3n+1).
$$
Now just expand both sides.
