I need to prove using induction that for $n \ge 4$ the inequality $n^3 < 3^n$ holds. I know that this problem has already been posted here but I managed to solve this in a simpler way than those offered by other proposers and so I decided to post it again to ask if it's actually correct.

(1) I check the basic case: $64 < 81 $ and so it works
(2) I assume that the statement is is true for all numbers greater than or equal to $k$
(3) for k+1: $$(k+1)^3=k^3+3k^2+3k+1 < k^3 + 3k^2 + 3k + k^3 < (*) $$ Now, I'd like to show that $3k^2 +3k < k^3 \Rightarrow k^2 - 3k -3 > 0 $Since this is an increasing parabola, this inequality will be satisfied for every $k$ greater than $n$, and $3 < n < 4 $ therefore, for $n = 4 n^3 > 3n^2 + 3n$
Therefore $$(*) < k^3 + k^3+k^3 = 3k^3 = 3(3^k) = 3^{k+1}$$ QED
What do you think of this solution?

  • $\begingroup$ In (1) you checked the statement for $n=4$. In (2) you assume the statement true for all $n$ at least 4 up to some $k$. (not "greater than or equal to k") In (3), you prove the induction step that if it is true for $k\geq4$ then the statement is also true for $k+1$. Your approach is then correct. (modulo the little edit in (2).) $\endgroup$ – Thibaut Dumont Nov 9 '17 at 22:56
  • $\begingroup$ You concept of "simpler" and mine vary drastically. $\endgroup$ – fleablood Nov 9 '17 at 23:04
  • $\begingroup$ last line $ 3k^3 < 3 (3^k) $ not $ 3k^3 = 3 (3^k) $ $\endgroup$ – user409387 Nov 9 '17 at 23:06
  • $\begingroup$ and $k^2−3k−3>0$ could be also shown with induction again. $\endgroup$ – user409387 Nov 9 '17 at 23:09

The middle part is a little weird. What you need to do is to show that if $k>4$ then $3k^2+3k < k^3$. From what you have written it looks as if you're going the other way ... and end up with $k < 4$ rather than starting with $k>4$ ... or was this supposed to be some kind of contraposition?

  • $\begingroup$ I agree that it was not originally obvious what I was trying to put across - I've edited my question. Thank you for your remark. Does it work now? $\endgroup$ – Aemilius Nov 9 '17 at 22:54

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