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.

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Are you sure it's true? $x_1=\frac{1}{2} \neq \frac{1}{\sqrt{10}}$ –  user1337 Mar 22 '13 at 14:50
What is the term in the product right before $\frac{2n-1}{2n}$? And right after $\frac{5}{6}$, actually... –  1015 Mar 22 '13 at 14:58
At the moment the simplest recursion works. Sooo... what did you try? –  Did Mar 22 '13 at 15:04
@julien: the term before $\frac{2n-1}{2n}$ is $\frac{2n-3}{2n-2}$ and after $\frac{5}{6}$ is $\frac{7}{8}$ –  A.D Mar 22 '13 at 15:15
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.