Doubt with a problem of mathematical induction

Question

I wanted to know if my solution for the following math induction problem is ok:

$\forall n ∈ \mathbb N \wedge n ≥ 4: 2^n ≥ n^2$

Is it well justified? In case this is wrong, I would like to know why

From already thank you very much.

Where have you proven that $2h^2\ge (h+1)^2$? I don't see any such proof. — Peter Foreman, Sep 10 '19 at 18:24
Peter, what I tried to show in that part is that, since n is a natural number> = 4, if the minimum case is met (4) it is true for the rest. Since they are all positive numbers Is the reasoning correct? — mndv, Sep 10 '19 at 18:30

score 1 · Answer 1 · answered Sep 10 '19 at 20:06

Prove: $\forall n \in \mathbb{N}, n\geq 4, 2^n \geq n^2 $

Base case: $n=4$, $2^4 \geq 4^2 \rightarrow 16 \geq 16. $

Inductive step:

Let's assume for $\forall n \in \mathbb{N}, n\geq 4,2^n \geq n^2$; Prove $2^{n+1} \geq {(n+1)}^2$

We know that $2^n\cdot 2 \geq 2n^2$

$(n-1)(n+1)>2n$ because $n+1 > n \ $ and $n-1>2$.

$n^2-1 > 2n$

$n^2 > 2n+1$

$2n^2 > n^2 + 2n+1$

$ 2n^2 > (n+1)^2$

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

Therefore, $2^n \geq n^2$ for $\forall n \in \mathbb{N}, n\geq 4 $ $\blacksquare$

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