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?

Let $S(n)$ be the statement: $3^{n}\geq{n^{2}}$; $n\geq{1}$
Basis step: $S(1)$:
LHS: $3^{(1)}=3$
RHS: $(1)^{2}=1$
$\hspace{77.5 mm}$ LHS $\geq$ RHS $\hspace{1 mm}$ (verified.)
Inductive step:
Assume $S(k)$ is true, i.e. assume that $3^{k}\geq{k^{2}}$; $k\geq{1}$
$\hspace{59 mm}\Rightarrow 3\bullet{3^{k}}\geq{3\bullet{k^{2}}}$
$S(k+1)$: $3^{k+1}$
$\hspace{12.5 mm}=3\bullet{3^{k}}$
$\hspace{11 mm}\Rightarrow 3\bullet{3^{k}}>3\bullet{k^{2}}$
$\hspace{11 mm}\Rightarrow 3^{k}+3^{k}+3^{k}>k^{2}+k^{2}+k^{2}$
$\hspace{40 mm}>k^{2}+2k+1=(k+1)^{2}$
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $3^{n}\geq{n^{2}}$; $n\geq{1}$.
