<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] Try expanding $2(k+1)^2 + 3(k+1)$ in order to compare it with $6k^2 + 9k$ (I.e., what you need to ensure that $6k^2 + 9k \geq 2(k+1)^2 + 3(k+1))$:
$$\begin{align} 2(k+1)^2 + 3(k+1) & = 2(k^2 + 2k + 1) + 3k + 3 \\ &= 2k^2 + 4k + 2 + 3k + 3 \\ &= 2k^2 + 7k +5\end{align}$$
Now, all that remains (while keeping in mind that $\bf k\geq 4)$ is to show that $$\begin{align} 6k^2 + 9k & \geq 6k^2 + 7k + 5 \\ 
& \geq 2k^2 + 7k + 5 \\ 
& = 2(k+1)^2 + 3(k+1)
\end{align}$$
So indeed, with the base case, and the fact that $P(k) \implies P(k+1)$, we have established that $P(n)$ holds for all $n \geq 4$.
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