How to prove this inequality $3^{n}\geq n^{2}$ for $n\geq 1$ with mathematical induction? Prove this inequality $3^{n}\geq n^{2}$ for  $n\geq 1$ with mathematical induction.
Base step: When $n=1$
$3^{1}\geq1^{2}$, statement is true. 
Inductive step: We need to prove that this statement $3^{n+1}\geq (n+1)^{2}$ is true. 
So, to get the left side of this statement is easy. We can get it by multiplying $3^{n}\geq n^{2}$ with $3$.
After this step we have $3^{n+1}\geq 3n^{2}$.
What we now have to get is the right side and we can transform it like this:
$3n^{2}=  (n^{2}+2n+1)+(2n^{2}-2n-1)$ which is same as
$(n+1)^{2}+(2n^{2}-2n-1)$.
So now we have $(n+1)^{2}$ and $(2n^{2}-2n-1)$ and my question is how should i use this to prove inequality?

$3^n\ge n^2, n\ge2$
$3^{n+1}=3^n+3^n+3^n\ge n^2+n^2+n^2\ge n^2+2n+1=(n+1)^2$ 
This uses the fact that $n^2\ge 2n,n\ge 2$ which can be shown via induction as well and obviously $n^2\ge1$ for all $n\ge 2$.  The case for $1$ can be shown and then use $2$ as the base case. 
