Can someone please check if my reasoning for this proof is valid or not?

Question

I have already seen the other questions about this proof. I'm just trying a different sort of method, though I'm not sure if it's valid or not.

Context for the main question: Prove by induction $2^n\gt n^3$ for $n\ge10$

Obviously, the base case works for n=10

$1024=2^{10}\gt1000=10^3$

The induction hypothesis: Assume $P_n$ is true $\rightarrow$ $2^n\gt n^3$

I want to then prove that $2^{n+1}\gt (n+1)^3$

Now, using the induction hypothesis:

$2^n\gt n^3$

multiply both sides by 2

$2^{n+1}\gt 2n^3$

Using the fact that $n\ge10$ this implies that $n^3\ge10n^2$

$2n^3=n^3 +n^3\gt n^3 +10n^2=n^3 +3n^2 +7n^2$

Using the fact that $7\gt 1$ this implies that $7n\gt n$ since n is positive.

$n^3 +3n^2 +7n^2\gt n^3 +3n^2 +n^2$

Once again, using $n\ge 10$ this implies $n^2\ge 10n$

$n^3 +3n^2 +n^2\gt n^3 +3n^2 +10n=n^3 +3n^2 +3n+7n$

Again $7\gt 1$

$n^3 +3n^2 +3n+7n\gt n^3 +3n^2 +3n +n$

Using $n\ge 10$ one last time

$n^3 +3n^2 +3n +n\gt n^3 +3n^2 +3n+10$

Since $10\gt 1$

$n^3 +3n^2 +3n+10\gt n^3 +3n^2 +3n+1=(n+1)^3$

Thus, through the chain of inequalities, I have proved that $2^{n+1}\gt (n+1)^3$.

QED

Sorry if there are any errors in my reasoning. Thank you for reading and feedback.

Answer 1

It is ok, but I think you can go faster.

If you want to prove that:

$$2n^3>(n+1)^3\Leftrightarrow 2>\left(\frac{n+1}{n}\right)^3=\left(1+\frac{1}{n}\right)^3$$

You can do:

$$n\ge10\to\frac{1}{n}\le0.1\to \left(1+\frac{1}{n}\right)^3\le (1.1)^3=1.331<2$$

Answer 2

Here's an alternate method: we have in our assumption that $$2^n>n^3,\quad n\geq 10$$

Then consider $f(n) = 2^{n+1}-(n+1)^3$

$f(n) = 2\cdot 2^n-n^3-3n^2-3n-1$

$f(n)= (2^n-n^3)+(2^n-3n^2-3n-1)$

Let $g(n) = n^3-3n^2-3n-1\implies g(n) <2^n-3n^2-3n-1$ using $2^n>n^3$

$g'(n) = 3n^2-6n-3 = 3(n^2-2n-1) =3[(n-1)^2-2] >0$ for $n\geq 10$

$g(10) = 693\implies g(n) >0,\quad n\geq 10$

Therefore $f(n)$ is the sum of two positive functions of $n$ $\implies f(n)>0,\quad n\geq 10$ as required

Answer 3

It's good but it'd be easier (both to do and follow) to go forward:

$(n+1)^3 = $$n^3 + 3n^2 + 3n + 1 $$< n^3 + 3n^2 + 3n^2 + 3n^2 = $$n^3 + 9n^2 < n^3 + n*n^2 = $$n^3 + n^3 =2n^3 $$< 2*2^n = 2^{n+1}$