Prove: $2^{n+2}\ge n^3$ induction proof

Question

Prove: $2^{n+2}\ge n^3 $ for every natural n

I got this exercise from TAU logarithms exercises, I tried to do the all usual induction technique but I got to a weird dead end:

I do the induction proof and I get to: $$2^{n+2}+2^{n+2} \ge n^3 +3n^2+3n+1$$

So $2^{n+2}\ge n^3$ and I try another induction proof and I get to:

$$2^{n+2}+2^{n+2} \ge 3n^2+3n+1+6n+6$$

So $2^{n+2}\ge 3n^2+3n+1$ and I try another induction proof but I get a wrong answer for the base case (which I think will be $n=2$ at this step) because $2^4$ is not greater than $18$

What am I doing wrong here?

If there are any other ways to solve this, it will also be welcomed.

Answer 1

Hint: Multiplying $$2^{n+2}\geq n^3$$ by $$2$$ we get

$$2^{n+3}\geq 2n^3$$ and you must show that $$2n^3>(n+1)^3$$

Answer 2

I'll just try to outline a way that popped into my head. As a base, for $n=0$, note that $2^2\geq 0$.

For the induction step, suppose $2^{n+2}\geq n^3$ for some $n$. Then $$2^{(n+1)+2}=2^{(n+2)+1}=2^{n+2}\cdot 2\geq n^3\cdot 2$$ by the induction hypothesis. Now, you just have to show that $2n^3\geq (n+1)^3$ from which then follows that

$$2^{(n+1)+2}\geq (n+1)^3$$

which is what you wanted to establish.

Answer 3

Going from n to n+1, the left side is multiplied by 2. The right side is multiplied by $((n+1)/n)^3$. The right side grows slower if $((n+1)/n)^3 < 2$, or $1+1/n < 2^{1/3} ≈ 1.26$, or $1/n < 0.26$, or $n > 1/0.26 ≈ 3.847$ or n ≥ 4.

So you prove the statement by hand for n = 1, 2, 3, 4. For n ≥ 4, $((n+1)/n)^3 < 2$, so $2^{n+2} > n^3$ implies $2^{n+3} > (n+1)^3$.