Inequality proof

  • Thread starter Panphobia
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  • #1
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Homework Statement


Use M.I. to prove that n! > n^3 for n > 5



The Attempt at a Solution



I already proved n! > n^2 for n>4, but this is nothing like that.
This is my inductive step so far.
n=k+1
(k+1)! > (k+1)^3

(k+1)! - (k+1)^3 > 0[/B]

(k+1)! - (k+1)^3 = (k+1)[k! - (k+1)^2]
> (k+1)[k^3 - (k+1)^2]

This is how far I can get, any hints? Am I going in the right track or am I going in the wrong direction completely?
 

Answers and Replies

  • #2
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5,764

Homework Statement


Use M.I. to prove that n! > n^3 for n > 5



The Attempt at a Solution



I already proved n! > n^2 for n>4, but this is nothing like that.
This is my inductive step so far.
n=k+1
(k+1)! > (k+1)^3[/B]
Don't start off with this - it's what you want to show. Use your assumption (i.e., that k! > k^3) and write (k + 1)! as (k + 1)k!
Panphobia said:

(k+1)! - (k+1)^3 > 0


(k+1)! - (k+1)^3 = (k+1)[k! - (k+1)^2]
> (k+1)[k^3 - (k+1)^2]

This is how far I can get, any hints? Am I going in the right track or am I going in the wrong direction completely?
 
  • #3
435
13
Don't start off with this - it's what you want to show. Use your assumption (i.e., that k! > k^3) and write (k + 1)! as (k + 1)k!
I only started it that way because I proved n! > n^2 the same exact way. Hmmm I am thinking about this at the moment.
 
  • #4
435
13
So then if k! > k^3, then (k+1)! > k^3(k+1), is this going on the right track, if it is I don't know what else to do. Wait, then do I have to prove that k^3(k+1) >= (k+1)^3 where k>5?
 
  • #5
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5,764
So then if k! > k^3, then (k+1)! > k^3(k+1), is this going on the right track,
No. You assume that k! > k^3, and then you use that assumption to show (prove) that (k + 1)! > (k + 1)^3.
Panphobia said:
if it is I don't know what else to do. Wait, then do I have to prove that k^3(k+1) >= (k+1)^3 where k>5?
Your work for that step should start off like this:
(k + 1)! = (k + 1) k! > (k + 1) k^3

Work with that last expression to show that it is larger than (k + 1)^3, at least when k > 5.
 
  • #6
435
13
I give up haha, I can't figure it out on my own, I know intuitively that (k+1)*k^3 > (k+1)^3, but don't know any way to prove it. I tried dividing out the (k+1) then leaving it as k^3 > (k+1)^2 then expanding (k+1)^2, but that was a dead end.
 
  • #7
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5,764
I give up haha, I can't figure it out on my own, I know intuitively that (k+1)*k^3 > (k+1)^3, but don't know any way to prove it. I tried dividing out the (k+1) then leaving it as k^3 > (k+1)^2 then expanding (k+1)^2, but that was a dead end.
You can't "divide out" anything, since you're not working with an equation or inequality. The last expression I showed in my previous post was (k + 1)k^3, which is the same as (equal to) k^4 + k^3. Intuitively, that should be larger than (k + 1)^3, being that the first is a 4th degree polynomial and the second is only a cubic.
 
  • #8
435
13
Yea, k^4 + k^3 > k^3 + 3k^2 + 3k + 1
k^4 > 3k^2 + 3k + 1

3k^2 + 3k + 1 < 3k^2 + 3k^2 + k^2 < 7k^2 < k^4

So k^3(k+1) > (k+1)^3
 
  • #9
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5,764
Yea, k^4 + k^3 > k^3 + 3k^2 + 3k + 1
Yeah, but you need to show why this is true, not just wave your arms. Possibly it might require a separate proof by induction.
Panphobia said:
k^4 > 3k^2 + 3k + 1

3k^2 + 3k + 1 < 3k^2 + 3k^2 + k^2 < 7k^2 < k^4

So k^3(k+1) > (k+1)^3
 
  • #10
435
13
Yea that's what I thought, is there no other simpler way to prove this?
 
  • #11
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5,764
k^4 > 3k^2 + 3k + 1 is equivalent to k^4 - 3k^2 - 3k - 1 > 0. It might be acceptable to show that the graph of y = x^4 - 3x^2 - 3x - 1 lies above the x-axis for all x larger than a specific x-value.
 
  • #12
Ray Vickson
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Yea, k^4 + k^3 > k^3 + 3k^2 + 3k + 1
k^4 > 3k^2 + 3k + 1

3k^2 + 3k + 1 < 3k^2 + 3k^2 + k^2 < 7k^2 < k^4

So k^3(k+1) > (k+1)^3
Do you really believe that 3k^2 + 3k^2 + k^2 < 7k^2? Saying things like that is a sure way to lose marks on an exam.
 
  • #13
435
13
Do you really believe that 3k^2 + 3k^2 + k^2 < 7k^2? Saying things like that is a sure way to lose marks on an exam.
On my sheet of paper its an equals sign, typo.
 

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