I'm given a inequality as such: $n! > n^3$ Where n > 5,
I've done this so far:
BC: n = 6, 6! > 720 (Works)
IH: let n = k, we have that: $k! > k^3$
IS: try n = k+1, (I'm told to only work from one side)
So I have (k+1)!, but I'm not sure where to go from here.
I've been told that writing out: $(k+1)! > (k+1)^3$ is a fallacy, because I can't sub in k+1
into both sides, but rather prove from one side only.
Any Help on how to continue from here, would be much appreciated.
You have $(k+1)!=(k+1)k!>(k+1)k^3$ by the induction hypothesis. Now it suffices to show that if $k>5$ then $k^3 > (k+1)^2$. To do that, we write $(k+1)^2 = k^2 +2k+1 < k^2 + kk + k^2 = 3k^2 < kk^2 = k^3$, since we're assuming $k>5$.