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?

Essentially, you want to show that $$3n^2 > (n+1)^2$$ which is not so hard since $$3n^2 - (n^2 + 2n + 1) > 0 \iff 2n^2 - 2n-1 > 0$$
But $2n^2 - 2n - 1 = 2(n^2 -n) - 1 = (n^2-2) + (n^2 - 2n+1) =(n^2 -2) + (n-1)^2$, so that we have for all $n \geq 2$ that $2n^2 - 2n - 1 \geq 0$ since $(n-1)^2$ is always $\geq 0$ and $n^2 - 2$ is $\geq 0$ when $n\geq 2$. Then this means that $$3n^2 \geq (n+1)^2$$ for all $n\geq 2$. And hence: 
$$3^n \geq n^2 \implies 3^{n+1} \geq 3n^2 \geq (n+1)^2$$
and we are done.

This is a very common tactic in inequality induction proofs, you have the inequality arising from your hypothesis; you then multiply both sides by something to get one side of the inequality in the required inductive form (say $a >b$, where $a$ is the required form) and then prove an inequality chain $b > \cdots > d$ where $d$ is the required form.
