Show that $2^{n} \geq (n +2)^{2}$ for all $n \geq 6$ Edit: If it is hard to read what I have written the essence of my question is: How come that $2 \times 2^{k} - (k+3)^{2} \geq 2^{k}$ from the assumption that $2^{k} \geq (k+2)^{2}$?

Show that $2^{n} \geq (n +2)^{2}$ for all $n \geq 6$ 
I have excluded steps:
Assumption:  $\textsf{LHS}_{k} \geq \textsf{RHS}_{k} = 2^{k} \geq (k+2)^{2}$
We want to show that $\textsf{LHS}_{k+1} - \textsf{RHS}_{k+1} \geq 0$
So I start as follows,
$\textsf{LHS}_{k+1} - \textsf{RHS}_{k+1} = 2^{k+1} - (k+3)^{2} = 2^{k} \times 2 - (k+3)^{2} = \textsf{LHS}_{k} \times 2 - (k+3)^{2} \geq \textsf{LHS}_{k} \geq \textsf{RHS}_{k}...$.  
(according to the assumption)
Here I need to stop because I do not understand how that is the case. 
I do not understand how $\textsf{LHS}_{k} \times 2 - (k+3)^{2} \geq \textsf{LHS}_{k}$ which is the same as $\textsf{LHS}_{k} \times 2 - \textsf{RHS}_{k+1} \geq \textsf{LHS}_{k}$  
I have no problem with $\textsf{LHS}_{k+1} > \textsf{LHS}_{k} \geq \text{RHS}_{k}$ nor $\textsf{LHS}_{k} \times 2 - \text{RHS}_{k}  \geq \textsf{LHS}_{k} \geq \text{RHS}_{k}$ it is the $\textsf{RHS}_{k+1}$ I have a problem with. 

If you have done the Base case and want to show that $\text{LHS}_{n+1} - \text{RHS}_{n+1} \geq 0$ you use the assumption that $2^{n} \geq (n+2)^{2}$ but also that $n \geq 6$.
In other words: 
$2^{n+1} - (n+3)^{2} = 2^{n} \times 2 - (n+3)^{2}$. 
Now due to the fact that $2^{n} \geq (n+2)^{2}$ it follows that if we substitute $(n+2)^{2}$ for $2^{n}$ that 
$2^{n+1} - (n+3)^{2} = 2\times 2^{n} - (n+3)^{2}$  $\geq 2 \times (n+2)^{2} - (n+3)^{2} = n^{2} + 2n - 1$ 
Use that $n \geq 6$  
$n^{2} + 2n - 1 = n\times n + 2\times n - 1$
$\geq  6\times n + 2 \times n - 1 $
$\geq 6\times 6 + 2\times n -1$
$ \geq 6 \times 6 + 2\times 6 - 1 $
$\geq 0$.  
