Prove that $3^n > 2n^2 + 3n$ for $n \in [4,\infty) \cap\mathbb{N}$ If $n$ is a natural number $n\ge 4$, prove that $3^n > 2n^2 + 3n$
I assume I am supposed to use induction. The base $n=4$ step is clear, but how do I prove the inductive step. I tried several things including comparing $f(n+1)/f(n)$ and $f(n+1)-f(n)$ but they didn't seem to help. A hint would be great!

Write
$$ f(n) = 3^n - 2 n^2 - 3 n$$
Then
$$ f(n+1) = 3^{n+1} - 2 \big(n+1\big)^2 - 3 \big(n+1) $$
which can be written as
$$ f(n+1) = 3 \big( 3^n - 2 n^2 -  3 n \big) + 6 n^2 + 9 n - 2 \big(n+1\big)^2 - 3\big(n+1)$$
So
$$ f(n+1) = 3 f(n) + 4 n^2 + 2 n - 5$$
Note that
$$ f(3) = 0 $$
And note that
$$ 4 n^2 + 2 n - 5 > 0 $$
for $n \ge 1$, whence
$$ f(n) > 0 $$
for $n \ge 4$,
therefore
$$ 3^n - 2 n^2 - 3 n > 0$$
for $n \ge 4$, or
$$ 3^n > 2 n^2 + 3 n$$
for $n \ge 4$.
