Verified answer

Proof by Mathematical Induction

Basic Step:

When n = 10, the inequality 2^n > n^3 is true

Inductive Step:

Assume n = k is true, where k ≥ 10, then

2^k > k^3 ................ (*)

Let g(k) = k^3 - 6k^2 - 2 ==> g'(k) = 3k^2 - 12k = 3k(k - 4) > 0 for k ≥ 10

And, g(10) = 398 > 0

Thus g(k) is strictly positive and increasing for k ≥ 10

==> k^3 - 6k^2 - 2 > 0

==> k^3 > 6k^2 + 2 ............. (**)

So, from (*)

2(2^k)> 2(k^3)

2(2^k) > k^3 + k^3

2(2^k) > (k - 1)^3 + (6k^2 + 2), because k^3 > (k-1)^3 and apply (**)

2^(k+1) > (k^3 - 3k^2 + 3k - 1) + (6k^2 + 2)

2^(k+1) > k^3 + 3k^2 + 3k + 1

2^(k+1) > (k + 1)^3

So n = k + 1 is also true

Hence by principles of Mathematical Induction, 2^n > n^3 for n ≥ 10

gôhpihán has a solid proof. I also had in mind to do a proof by induction, but without using derivatives:

Same start: 2¹⁰ = 1024 > 1000 = 10³

Now, if n = k ≥ 10, then

2 > 1.331 = (10+1)³/10³ ≥ (k+1)³/k³

[ Because (10+1)³/10³ = 1 + 3/10 + 3/100 + 1/1000 ≥ 1 + 3/k + 3/k² + 1/k³ = (k+1)³/k³ ]

and so, from

2^k > k³

we can conclude that

2•2^k > k³•(k+1)³/k³

2^(k+1) > (k+1)³

QED

EDIT:

gôhpihán's proof can also be done without the use of derivatives, by establishing

g(k+1) > g(k), for k ≥ 10

algebraically. And it's a neat trick he's used, worthy of some study!

What the he'll does that say?