For every integer $n \geq 1$, prove that $3^n \geq n^2$. It's been a while since I've done induction, and I feel like I'm missing something really simple. What I have is this:

Base Case: $n=1$
   $$3^n \geq n^2 \implies 3 \geq 1$$
Inductive Hypothesis 
For all integers $1 \leq n < n+1$:
  $$3^n \ge n^2$$
Inductive Step 
$$3^{n+1} \geq \left( n+1\right) ^2 \implies 3\cdot 3^n \geq n^2+2n+2$$

This is as far as I got. Is it acceptable to say $2\cdot 3^n \geq 2n+2$ and then prove that? Or is there an easier way?

For $n>1$ we have $2n(n-1)\geq 1$, thus:
$3^n\geq n^2\rightarrow 3^{n+1}\geq 3n^2=n^2+2n^2$
According to the first line of the answer, $2n(n-1)\geq 1\rightarrow 2n^2\geq 2n+1$
So we can say:
$3^{n+1}\geq n^2+2n+1=(n+1)^2$
Hence the assumption is proved by induction
