Standard inductive problem

Question

Question: Prove that $2^n \geq (n+1)^2$ for all $n \geq 6$.

I have tried to prove this below and I'm interested if my method was correct and if there is a simpler answer since my answer seems unnecessarily long for such a simple claim.

Inductive hypothesis $$2^n \geq (n+1)^2$$

We need to show that $2^{n+1} \geq (n+2)^2$ or alternatively $$2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$$

claim $$\frac{(n+2)^2}{(n+1)^2} <2, \forall n \geq 6$$

notice that it is true for $n=6$ and $$\frac{\frac{(n+2)^2}{(n+1)^2}}{\frac{((n+1)+2)^2}{((n+1)+1)^2}}<1$$ so $\frac{(n+2)^2}{(n+1)^2}$ is decreasing as $n$ gets larger so we have proven the above claim.

So we have that for all $n \geq 6$ we have $2^n \geq (n+1)^2$ and by our induction hypothesis and $\frac{(n+2)^2}{(n+1)^2} <2$ so we can conclude that $2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$.

Answer 1

Except for showing a base case I think your method is good. Note that without a base case you do not have an induction, so make sure you pick e.g., $n=6$ and note that $2^6=64\ge (6+1)^2=49.$ You might consider trying to show the inductive hypothesis by showing that $n\ge 6\implies (n+1)^2\ge 2n+3$ and thus

$$2\cdot 2^n\ge 2(n+1)^2 \ge (n+1)^2+2n+3=(n+2)^2$$

where $2n+3$ is the difference between $(n+1)^2$ and $(n+2)^2$.

Answer 2

You have $2^{n+1}=2\times 2^{n}\geq 2(n+1)^2$ Now, for $n\geq 6,$ $$2(n+1)^2-(n+2)^2$$ $$=2(n^2+2n+1)-(n^2+4+4n)$$ $$=2n^2+4n+2-n^2-4-4n$$ $$=n^2-2\geq0$$ $$\therefore 2(n+1)^2\geq(n+2)^2$$ $$\therefore 2^{n+1}\geq(n+2)^2$$

:)