<s>[INST] Proof by induction of inequality. For $n \geq 2, 3^n \gt 2^n+n^2$ I'm trying to understand proof by induction on inequalities.
One problem I have is:

Prove that for $n \geq 2$  $3^n \gt 2^n+n^2$

I can prove the base case as $n=2$:

$3^2 \gt 2^2+2^2$ 
$9 \gt 8$

What I don't understand is the inductive step. How do I show for some $k$ where $k \geq 2$, $3^{k+1} \gt 2^{k+1} + (k+1)^2$.
 [/INST] Suppose $3^k > 2^k + k^2$ you have to show $3^{k+1} > 2^{k+1} + (k+1)^2$. From the inductive step, $3^{k+1} = 3\cdot 3^k > 3(2^k+k^2)= 2^{k+1}+2^k+k^2+2k^2>2^{k+1}+k^2+2k+1=2^{k+1}+(k+1)^2$. This is true because $2^k \ge 2k$, and $2k^2 > 1$. Thus by induction, we are done.
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