Can someone shed some light on this inequality? I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$\frac{a_{n+1}}{a_n}>\left (1-\frac{1}{n+1}\right ) \left (\frac{n+1}{n}\right)$$
where does the equation in the first and second parenthesis come from? 
Ok, I have another relating question:
why $$\frac{a_{n+1}}{a_n}> (1+\frac{1}{n})$$ ( The expression of third line.

 A: From Bernoulli's inequality, we have 
$$\left( 1- \frac{1}{(n+1)^2}\right) > 1+(n+1) \left(\frac{-1}{(n+1)^2} \right)=1-\frac1{n+1}$$
Hence, 
$$\frac{a_{n+1}}{a_n}>\left( 1- \frac{1}{(n+1)^2}\right)\left( \frac{n+1}{n}\right)>\left(1-\frac1{n+1} \right)\left( \frac{n+1}{n}\right)$$
A: It is putting together the result from the first red box with the second one:


*

*$\frac{a_{n+1}}{a_n} = \color{blue}{\left(1- \frac{1}{(n+1)^2} \right)^{n+1}}\left( \frac{n+1}{n}\right)$

*$\color{blue}{\left(1- \frac{1}{(n+1)^2} \right)^{n+1}} > \color{green}{1 + (n+1)\left( \frac{-1}{(n+1)^2}\right)}$
$$\Rightarrow \frac{a_{n+1}}{a_n} > \left(\color{green}{1 + (n+1)\left( \frac{-1}{(n+1)^2}\right)}\right)\left( \frac{n+1}{n}\right) = \left(\underbrace{1- \frac{1}{n+1}}_{=\frac{n}{n+1}}\right)\left( \frac{n+1}{n}\right)$$
A: So, we have
$$\frac{a_{n+1}}{a_n} = \left(1 - \frac{1}{(n+1)^2}\right)^{n+1}\left(\frac{n+1}{n}\right).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$\left(1 - \frac{1}{(n+1)^2}\right)^{n+1} > 1 + (n+1)\left(\frac{-1}{(n+1)^2}\right) = 1 - \frac{1}{n+1}.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$\frac{a_{n+1}}{a_n} = \left(1 - \frac{1}{(n+1)^2}\right)^{n+1}\left(\frac{n+1}{n}\right) > \left(1-\frac{1}{n+1}\right)\left(\frac{n+1}{n}\right).$$
Finally, we multiply out the RHS of the inequality 
$$\frac{a_{n+1}}{a_n} > \left(1-\frac{1}{n+1}\right)\left(\frac{n+1}{n}\right) = \frac{n+1}{n} - \frac{1}{n} = 1.$$
So, we have
$$\frac{a_{n+1}}{a_n} > 1 \implies a_{n+1} > a_n,$$
which means that $\{a_n\}$ is an increasing sequence. 
