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$$2^n \ge n^3 \ \forall n\ge 1$$ My effort:

  1. Check the base case: $P(1): 2\ge1 \ \text{(true)}$
  2. supposing that $P(n) \Rightarrow P(n+1):$ $2^{n+1} \ge (n+1)^3$
  3. I'll try the induction step: $$2^{n+1} = 2^n \cdot 2 \ge 2n^3 = n^3 \cdot 2 \ge n^3 \cdot \frac{1}{(n+1)^3}$$ I'm not sure about the last step, but in any case, I can't find any way to get to $(n+1)^3$ and prove the induction step.

How can I do?

José Carlos Santos
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Loris Simonetti
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    What is the idea behind dividing by $(n+1)^3$ in your last step? As it was noted below, you can prove this for $n\geq 10$. Just simply show that $2n^3 \ge (n + 1)^3$ for $n\geq 10$. – Gary Apr 01 '21 at 14:36

1 Answers1

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You cannot prove that since it is false when $n=2$. It is true when $n\geqslant10$ though. Indeed, $2^{10}=1024>1000=10^3$, and, if $2^n\geqslant n^3$, then, as you have proved, $2^{n+1}\geqslant2n^3$. And $2n^3\geqslant(n+1)^3$, since$$2n^3\geqslant(n+1)^3\iff2\geqslant\left(1+\frac1n\right)^3,$$which is true, since $n\geqslant10$, and therefore$$\left(1+\frac1n\right)^3\leqslant\left(\frac{11}{10}\right)^2<2.$$

José Carlos Santos
  • 427,504