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?
 A: Sort of. I prefer to do it in a single chain of inequalities though.

Inductive Hypothesis: Assume that the desired inequality holds for all $n' \in \{1, \ldots, n\}$, where $n \geq 2$ (you'll need to do an extra base case for $n = 2$).
It remains to prove that the desired inequality holds for $n' = n + 1$. Indeed, observe that:
\begin{align*}
3^{n+1}
&= 3(3^n) \\
&\geq 3(n^2) &\text{by the inductive hypothesis}\\
&= n^2 + (n)n + (n)^2 \\
&\geq n^2 + (2)n + (n)^2 &\text{since }n \geq 2\\
&> n^2 + (2)n + (1)^2 &\text{since }n \geq 2 > 1\\
&= (n + 1)^2
\end{align*}
as desired.
A: 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
A: The claim is clear if $n=1$.
If $n\ge 2$, by the binomial theorem, we have
$$
3^n = (1+2)^n = 1 + 2\binom n1 + 2^2\binom n2 + \dots > 2\binom n1 + 2^2\binom n2 = 2n + 2n(n-1) = 2n^2 > n^2
$$
