Proof By Induction $n^2 > 3n$ where $n\ge 4$ I am trying to prove the following example, however I seem to be getting a little stuck:
For $n\in\mathbb N$, $n\ge 4, n^2>3n$
What I have Done:
Base Case:$ n=4$, LHS: $4^2 = 16$, RHS: $3\cdot 4 = 12$
$16\gt 12$, so True
Assume true for $n=k$,
$k^2 > 3k$
Should be true for $n=k+1$
$(k+1)^2 \gt 3(k+1)$
This is where I am stuck!
Any help would be appreciated!
 A: $(k+1)^2=k^2+2k+1>3k+2k+1>3k+3=3(k+1)$
A: Paul's answer is perfectly fine, but it seems you may still be struggling with how exactly he coaxed out the inequality you are looking for by using the induction hypothesis. I'll try to expand on this--let me know if a step(s) doesn't make sense. 

First, for $n\geq 4$, let $S(n)$ denote the statement
$$
S(n) : n^2 > 3n.
$$
You already took care of the base case. Thus, fix some $k\geq 4$ and assume that $S(k)$ is true where
$$
S(k) : \color{blue}{k^2 > 3k}.
$$
To be shown is that $S(k+1)$ follows where 
$$
S(k+1) : \color{purple}{(k+1)^2 > 3(k+1)}.
$$
Starting with the left-hand side of $S(k+1)$,
\begin{align}
\color{purple}{(k+1)^2} &= \color{blue}{k^2}+2k+1\tag{expand}\\[0.5em]
& \,\color{blue}{>3k}+2k+1\tag{by $S(k)$, the ind. hyp.}\\[0.5em]
&> 3k+3\tag{since $k\geq 4$}\\[0.5em]
& \,\,\color{purple}{>3(k+1)},\tag{factor out $3$}
\end{align}
we end up at the right-hand side of $S(k+1)$, completing the inductive step.
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 4$. $\blacksquare$
