prove that $6^n \ge n3^n$ by induction

Question

I am trying to prove that $6^n \ge n3^n$ for all $n\ge 0$ but I am getting stuck at solving the hypothesis I have proven that at the base case of 0 and 1 the inequality is true.

With that in mind I assumed that n=k will keep the inequality of $6^k\ge k3^k$

So in the hypothesis I am trying to show that $n=k+1$ starting with my assumption I have $6^k \ge k3^k$

$6*6^k \ge (6)k3^k$ multiply both sides by 6 to make it $6^{k+1}$

$6^{k+1} \ge (6)k3^k$ but here is where I get stuck
How can I make the right hand look like my hypothesis

Answer 1

$$ (6)k3^k =2k \cdot 3 \cdot 3^k=(k+k) \cdot 3^{k+1} \geq (k+1)\cdot 3^{k+1}$$

Answer 2

It's actually much simpler than it sounds. Both sides have a common factor of $3^n$, so after division, we get $2^n => n$.

If $n$ is negative, $2^n$ is greater, as it will always equal a positive (but tiny) value.

If $n$ is zero, $2^n$ is greater, since 2^0 = 1, which is greater than 0.

If $n$ is positive, $2^n$ is greater. $2^n$ grows at a faster rate than $n$, and is already greater at 0, so the gap just continues to widen.