Proof by induction that $3^{n+1} > n^4$

Question

Prove $3^{n+1} > n^4$ for all $n \in \mathbb{N}$, $n \neq 3, n \neq 4$.

Let P(n) be the statement "$3^{n+1} > n^4$ such that $n \in \mathbb{N}$, $n \neq 3, n \neq 4$."

I have proved the base cases, P(1), P(2), P(5), but I got stuck on the inductive step.

I said: Let $k \in \mathbb{N}, k \geq 5$, assume P(k) is true. Show P(k+1) is true.

$P(k+1) = 3^{k+2} > (k+1)^4 $

I got stuck here and don't know how to proceed, any help is greatly appreaciated!

Answer 1

Hint: given $P(k)$, that is, $3^{k+1} > k^4$, you want to show $P(k+1)$, which is

$$3^{k+2} > (k+1)^4$$

You might try

$$3^{k+2} = 3 \times 3^{k+1} > 3 k^4\ \ \ \ \text{(induction hypothesis)}$$

and see if this is bigger than $(k+1)^4$.