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. 
 A: Hint. The induction step could be done more simply. Note that
$$\left(\frac{n+3}{n+2}\right)^2\leq 2,~\forall n\geq 6$$
A: Assume that $2^k\geq (k+2)^2$.
Then
$\begin{align*}
2^{k+1}&=2(2^k)\\
&\geq 2(k+2)^2\\
&=2(k^2+4k+4)\\
&=(k^2+6k+9)+k^2+2k-1\\
&=(k+3)^2+k^2+2k-1\\
&\geq (k+3)^2
\end{align*}$
since $k^2\geq 0$ and $2k-1\geq 0$ for $k\geq 6$.
A: Base Case:
$$2^{6} =64\geq (6 +2)^{2}=64$$
Inductive Step:
Assume true for some $k\geq 6$
$$2^{k} \geq (k +2)^{2}$$
Now show true for $n=k+1$ from assumed truth of $n=k$ case.
$$2^{k+1} \geq ((k+1) +2)^{2}=k^2+6k+9$$
so
\begin{align*}
2^{k+1}&=2^k\cdot 2 \geq 2(k+2)^{2}\\
&=2k^2+8k+8\geq ((k+1) +2)^{2}
\end{align*}
for $k\ge6$.
A: 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 $
$\geq 0$.  
