A problem of inequality and request for help Problem
If $n$ be a positive integer, prove that 
$$\begin{align}
\frac{1}{2\sqrt{n+1}} < \frac{1.3.5 \dots (2n-1)}{2.4.6 \dots 2n}< \frac{1}{\sqrt{2n+1}}
\end{align}$$

Following is my steps to solve the above problem
Taking AM>GM on $1.3.5 \dots (2n−1)$ and $2.4.6 \dots 2n$, separately, we get
  $1.3.5 \dots (2n−1)<(1+3+5+ \dots (2n-1))^{n-1}=(n(n-1))^{n-1}$
$2.4.6 \dots 2n<(2+4+6+ \dots 2n)^{n}=(n(n+1))^{n}$ 
Then I tried to prove it but not get the answer. The calculation that I have done is not fruitful.

 A: In inductive reasoning, you prove that if something is correct for $n$, then it is correct for $n+1$ too, hence for all number. Of course, it must be correct for 1.
consider the middle value as $a_n$
Then $$a_{n+1}=a_n \times \frac{2n+1}{2n+2}$$
You need to prove
$$\frac{1}{2 \sqrt{(n+1)+1}}<a_{n+1}<\frac{1}{\sqrt{2(n+1)+1}}$$
Here is how to do that:
$$a_{n+1}=a_{n} \times \frac{2n+1}{2n+2} ~~~(I)$$
According to the question
$$a_{n}<\frac{1}{\sqrt{2n+1}} ~~~(II)$$
combine (I) and (II):
$$a_{n+1}<\frac{1}{\sqrt{2n+1}} \times \frac{2n+1}{2n+2}=\frac{\sqrt{2n+1}}{2n+2}<\frac{\sqrt{2n+1}}{2n+1}\stackrel{?}{<}\frac{1}{\sqrt{2n+3}}$$
This proves the right inequality. Thes same must be done for the left equality. You may ask, how did I conclude the last part:
$$\frac{\sqrt{2n+1}}{2n+1}\stackrel{?}{<}\frac{1}{\sqrt{2n+3}}$$
if and only if
$$\sqrt{2n+1} \times \sqrt{2n+3} < \sqrt{2n+2} \times \sqrt{2n+1}$$
if and only if
$$(2n+1)(2n+3)<(2n+2)(2n+2)$$
Which is always correct for natural numbers.
To prove the left inequality, you will reach to 
$$\frac{1}{2\sqrt{n+2}}\stackrel{?}{<}\frac{2n+1}{2n+2}\times \frac{1}{2 \sqrt{n+1}}$$
if and only if
$$\frac{1}{\sqrt{n+2}}\stackrel{?}{<}\frac{2n+1}{2n+2}\times \frac{1}{ \sqrt{n+1}}$$
if and only if
$$\frac{\sqrt{n+1}}{\sqrt{n+2}}\stackrel{?}{<}\frac{2n+1}{2n+2}$$
if and only if
$$(n+1)(2n+2)^2<(n+2)(2n+1)^2$$
