Inequality and Induction: $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$ I needed to prove that   
$\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$,  $\forall n \geq 1$ .    
I've atempted by induction.   
I proved the case for $n=1$ and assumed it holds for some $n$.
The left-side of the n+1 case is
$\frac{1}{2}.\frac{3}{4}. ...  .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2}$.   
Using the inductive hypothesis, i could reach the result that 
$\frac{1}{2}.\frac{3}{4}. ...  .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} <\frac{2n+1}{\sqrt{2n+1}(2n+2)}=\frac{\sqrt{2n+1}}{(2n+2)}$.    
Now, i'm wondering how should i connect it to my goal :   
$\frac{1}{\sqrt{2n+2+1}}=\frac{1}{\sqrt{2n+3}} $    
I know one way to prove that 
$\frac{\sqrt{2n+1}}{(2n+2)}<\frac{1}{\sqrt{2n+3}}  $   
We just square things, then eventually reach   
$ (n+1).(n+3)<(n+2)^2$   
Which is easily provable because  $3<4$ ....
But I was wondering if there was another way to show that ... perharps a more direct way to show that last bit ... a way that was not so direct and brute as to involve squaring both sides.  A way of gradually manipulating the left-side until reaching the inequality with the right side.   
Thanks in advance.

$$\begin{align}\frac{1}{2}.\frac{3}{4}. ...  .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} &<\frac{2n+1}{\sqrt{2n+1}(2n+2)}\\~\\&=\frac{\sqrt{2n+1}}{(2n+2)}\\~\\&=\frac{\sqrt{(2n+1)(2n+3)}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{\sqrt{4n^2+8n+3}}{(2n+2)\sqrt{2n+3}}\\~\\&\lt \frac{\sqrt{4n^2+8n+3+\color{blue}{1}}}{(2n+2)\sqrt{2n+3}}\\~\\&= \frac{\sqrt{(2n+2)^2}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{1}{\sqrt{2n+3}}\end{align}$$

Alternatively you may also use AM-GM inequality and conclude directly that $\sqrt{(2n+1)(2n+3)}\lt  2n+2$
