Proving 3n > n^3 for n > 3 using Induction

[rbnphlp]

3ⁿ>n³ where n >3

I know I have to use proof by induction to solve this.
assume for f(4)is true
for f(n+1)=3ⁿ⁺¹>3n³

However after that I don't have a clue of how to getting it into (n+1)³
Any hints will be greatly apperciated

 

[statdad]

Let

$$A_n$$

be the statement

$$3^n > n^3$$

You want to show $$A_n$$ is true for $$n \ge 4$$.

It shouldn't be hard to show $$A_4$$ is true. Assume it is true for $$k \ge 4$$.

Now

$$3^{3+1} = 3 \cdot 3 ^k > 3 \cdot k^3 = k^3 + k^3 + k^3$$

Now play with the terms on the right, making the new expressions ever smaller, to build up to the expansion of $$(k+1)^3$$.

(Hint for a start: You know by hypothesis $$k \ge 4 > 3$$, so
$$k^3 = k \cdot k^2 > 3k^2$$
which is the second term in the expansion of $$(k+1)^3$$)

 

[rbnphlp]

Let

$$A_n$$

be the statement

$$3^n > n^3$$

You want to show $$A_n$$ is true for $$n \ge 4$$.

It shouldn't be hard to show $$A_4$$ is true. Assume it is true for $$k \ge 4$$.

Now

$$3^{3+1} = 3 \cdot 3 ^k > 3 \cdot k^3 = k^3 + k^3 + k^3$$

Now play with the terms on the right, making the new expressions ever smaller, to build up to the expansion of $$(k+1)^3$$.

(Hint for a start: You know by hypothesis $$k \ge 4 > 3$$, so
$$k^3 = k \cdot k^2 > 3k^2$$
which is the second term in the expansion of $$(k+1)^3$$)

Sorry for the late reply
This is th best I could get at:

3ⁿ>n³
For n+1=
3ⁿ⁺¹>3n

given n>3 , I found the following:

• n³>3n²

• n²>3n

• n-2>1

with the best intentions I did the following:

Added all them up and got
n³>3n²+3n+1-n²-n+2--(1)
3ⁿ⁺¹-2n³>n³--(2)

from 1 and 2
3ⁿ⁺¹-2n³>3n²+3n+1-n²-n+2

3ⁿ⁺¹+n²+n-2-n³>(n+1)³

But after that I am stuck it would be great some one could help. thnx

 

[rbnphlp]

anyone?

 

[HallsofIvy]

Statdad showed that you could get to $k^3+ k^3+ k^3$ and you want to compare that to $(k+1)^3= k^3+ 3k^2+ 3k+ 1$. Can you show that $k^3> 3k^2$ when k> 3? Can you show that $k^3> 3k+ 1$ when k> 3?

 

[rbnphlp]

Statdad showed that you could get to $k^3+ k^3+ k^3$ and you want to compare that to $(k+1)^3= k^3+ 3k^2+ 3k+ 1$. Can you show that $k^3> 3k^2$ when k> 3? Can you show that $k^3> 3k+ 1$ when k> 3?

nope I can't .