A question on Induction

Question

Prove that: for any natural number $n\ge10$, $$2^n\ge n^3.$$ How can I prove by induction if $n\ge10$ and I must do that $2^n\ge n^3$?

I stop on this step: $$2^{n+1}\ge n^3+3n^2+3n+1.$$

Here's the question I'm trying to prove. I'm just not certain how I should approach the inductive / constructor step. I proved induction wrote $n+1$ instead of $n$.

I think I have everything right until the induction step. But I don’t know what I should do next step, because I had never solve inequality before I began ask questions on this site.

Answer 1

HINT:

The basis for induction is easy (just substitute $n=10$).

For the induction hypothesis, suppose that $n=k$ holds, so $$2^k\ge k^3$$ for some $k\ge 10$.

Consider $n=k+1$:

$$2^{k+1}=2\cdot2^k\ge2k^3$$ and all you have to do now is show that $$2k^3\ge(k+1)^3$$ for $k\ge10$.

Answer 2

$2^(n+1) = 2^n*2 >= 2n^3$

Compare $2n^3$ to $n^3 + 3n^2 + 3n + 1$

$n^3 vs. 3n^2 + 3n + 1$

We can start to see that the left hand side is greater than the right hand side for n>=10. It is true at n=10.

Consider each side as two functions. Then taking derivatives we have $3n^2$ and $6n + 3$

The left function is increasing faster than the right function for n > 10.