Proof that \forall n \in \mathbb{N}: 3^{n} \geq n^{3}

[Moridin]

Homework Statement

Show that

$$\forall n \in \mathbb{N}: 3^{n} \geq n^{3}$$

The Attempt at a Solution

(1) Show that it is true for n = 1:

$$3^{1} \geq 1^{3}$$

(2) Show that if it is true for n = p, then it is true for n = p + 1:

Assume that $$3^{p} \geq p^{3}$$

Now,

$$3^{p+1} = 3 \cdot 3^{p} = 3^{p} + 3^{p} + 3^{p}$$

$$(p+1)^{3} = p^{3} + 3p^{2} + 3p + 1$$

Given our assumption, we know that if it could be demonstrated that

$$3^{p} + 3^{p} \geq 3p^{2} + 3p + 1$$

then we are done. From here, I'm not sure how to proceed. Should I pull some moves from analysis and argue that certain functions grow faster than others above a certain n? The last inequality is also a stronger criteria, but does not apply to p = 1 or p = 2, since 3^{p} was larger than p^{3}.

 

[dynamicsolo]

Why not just compare

$$3^{p} + 3^{p} + 3^{p}$$

term-by-term with

$$p^{3} + 3p^{2} + (3p + 1)$$ ?

You've assumed

$$3^{p} \geq p^{3}$$

and, beyond some low value of p,

$$3^{p} \geq 3p^{2}$$

and

$$3^{p} \geq 3p+1$$.

EDIT: the last two inequalities only fail for p = 1, but you've already shown that the proposed inequality works there. As for showing that the inequalities work for p>= 2,
how about taking log base 3 of both sides?

 

[Architeuthis Dux]

Your attempt at a solution is a good start. To proceed, you can use mathematical induction to prove that the statement holds true for all natural numbers.

First, we have already shown that the statement is true for n = 1.

Next, we assume that the statement is true for some arbitrary natural number k, i.e. 3^{k} \geq k^{3}.

Now, we need to show that the statement is also true for k+1.

Using the assumption, we have 3^{k+1} = 3 \cdot 3^{k} \geq 3k^{3} (since 3^{k} \geq k^{3}).

We also have (k+1)^{3} = k^{3} + 3k^{2} + 3k + 1.

Since k^{3} \geq k^{3}, 3k^{3} \geq 3k^{2} and 3k^{3} \geq 3k, we can conclude that 3^{k+1} \geq (k+1)^{3}.

Therefore, by mathematical induction, we have shown that the statement is true for all natural numbers.

In conclusion, we have proven that \forall n \in \mathbb{N}: 3^{n} \geq n^{3}.