Verify Proof by Mathematical Induction: $n^2 > 4n+1$ I am just learning proof by mathematical induction and wanted to verify if I got the following proof right

Use induction to prove $n^2 > 4n + 1$

Proceed with induction. For $n = 5$. The left hand side of is 25 and the right hand side is 21. Therefore the claim is valid for $n = 5$.
Now, assume that the claim is valid for $n = k$ where $k$ is some integer $ \geq 5$. That is,
$$ k^2 > 4k +1$$
Now, for $k \geq 5$ where $k$ where $k$ is an integer, prove that the claim is valid for $n = k+1$. That is,
$$(k+1)^2 > 4k +5$$
We being with our assumption that
$$ k^2 > 4k +1$$
$$ k^2 + 4 > 4k +5$$
We know that $(k+1)^2 = k^2 +2k +1 > k^2 +4$ for $k \geq 5$. Therefore we can substitute $(k+1)^2$ for $k^2 +4$
And, $$(k+1)^2 > 4k + 5$$
By the principle of mathematical induction, the claim is valid for $n \geq 5$
I also have another alternate way of proving but didn't know if one is better than the other or both are wrong. This would be in the proving step
$(k+1)^2 = k^2+2k+1 > 4k+1+2k+1 = 6k +2 > 4k +5$ for $k \geq5$
 A: Your proof is ultimately correct but could be cleaned up a good bit--I'd recommend reading this post about how to write a clear induction proof. Here is how I would write the core part of your induction proof:
$$
\begin{align}
(k+1)^2
&= k^2+2k+1 & \text{(expand)}\\[0.5em]
&> (4k+1)+2k+1 & \text{(by induction hypothesis)}\\[0.5em]
&= [4(k+1)+1]+(2k-3) & \text{(rewrite)}\\[0.5em]
&> 4(k+1)+1. & \text{(since $2k-3>0$ when $k\geq5$)}
\end{align}
$$
It is clear from this problem that you are learning the mechanics of proofs by induction, but hopefully it is also clear that this is a problem that really does not require a proof by induction. In fact, a basic application of the quadratic formula shows that the zeros of the quadratic $k^2-4k-1$ are $k=2\pm\sqrt{5}$. Since you are trying to show that $n^2>4n+1$ for $n\geq5$, where $n$ is an integer, your problem boils down to trying to show that $n^2-4n-1>0$, and the quadratic formula indicates this is the case when 
$$
n\in(-\infty,2-\sqrt{5})\cup(2+\sqrt{5},\infty),
$$
or, restricting our attention to integers, 
$$
n\in(-\infty,\lfloor2-\sqrt{5}\rfloor]\cup[\lceil2+\sqrt{5}\rceil,\infty)=(-\infty,-1]\cup[5,\infty).
$$
The point is that induction is not at all necessary in the course of working out the validity of your inequality. But learning how to write a good induction proof is good practice nonetheless.
A: Yes, the proof is correct. But this result can be obtained in a much more elementary manner, if you're not restricted to using induction. Observe that the inequality to be proved is the same as $n^2-4n-1>0$, or $(n-2)^2>5$. This is obvious when $n$ is a real number either smaller than $\sqrt5-2$, or larger than $\sqrt5+2$. In particular, when $n$ is an integer, then every $n\geq5$ or $n\leq-1$ satisfies the inequality.
