<s>[INST] Proof by induction that $3^n \geq 2n^2 + 3n$ for $n \ge 4$ Problem:
If $n$ is a natural number and $n\geq4$, then $3^n \geq 2n^2 + 3n$. (Prove by Induction.)
Attempt at solution:
1) Given: $n$ is a natural number, $n \geq 4$.
2) Let $P(n)$ be the statement "$3^n \geq 2n^2 + 3n$."
3) $P(4) = 3^4 > 2(4)^2 + 3(4)$, i.e. $81 > 44$, thus the base case of $P(4)$ is true.
4) Assume $P(k) = 3^k \geq 2k^2 + 3k$ is true, where $k$ is a natural number and $k \geq 4$.
5) $P(k + 1) = 3^k(3) = 3^{k+1} > (2k^2 + 3k)(3) = 6k^2 + 9k$.  
And this is where I am stuck. I know that I am trying to get $P(k+1)$ into the form $3^{k+1} > 2(k+1)^2 + 3(k + 1)$, and I have tried all sorts of algebraic manipulation, but I still am nowhere close to arriving at the correct form. How should I proceed?
 [/INST] $3^n\geq2n^2+3n$
(1) $n=4$ 
$81\geq44$
(2) Assume true for $n=k$ ,Therfore we have :
$3^k\geq2k^2+3k$
(3) Then for $n=k+1$
$3^{k+1}\geq2(k+1)^2+3k+3$
Simplifying this gives: 
$3(3^k)\geq2k^2+7k+5$
Now to prove this ^^^ you can subsititute what we assumed in (2) to show this inequality stands.
$3^k\geq2k^2+3k$
So..
$3(2k^2+3k)\geq2k^2+7k+5$ 
$6k^2+9k\geq2k^2+7k+5$ 
$6k^2+7k+2k\geq2k^2+7k+5$ 
^^^ This is true, therefore its true for n=k+1 
So there you have it
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