Can AM-GM inequality prove 2^n > 1+n\sqrt{2^{n-1}} for positive integers?

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Homework Help Overview

The discussion revolves around proving the inequality \(2^n > 1 + n\sqrt{2^{n-1}}\) for positive integers \(n\). Participants are exploring the application of the AM-GM inequality and other mathematical techniques to approach this problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Some participants consider using the AM-GM inequality but are uncertain about the appropriate numbers to apply it to. Others question the validity of the inequality for specific values of \(n\), noting that for \(n=1\), the two sides are equivalent. There is also mention of using mathematical induction and the binomial theorem as potential methods to prove the inequality.

Discussion Status

The discussion is ongoing, with participants sharing various approaches and expressing uncertainty about the best method to use. Some guidance has been offered regarding the use of mathematical induction and the binomial theorem, but no consensus has been reached on a definitive approach.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to prove the inequality for all positive integers \(n\) and the implications of using different mathematical techniques.

utkarshakash
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Homework Statement


If n is a positive integer, prove that [itex]2^n > 1+n\sqrt{2^{n-1}}[/itex]

Homework Equations



The Attempt at a Solution


I am thinking of applying AM GM HM inequality. But which numbers should I take to arrive at this inequality?
 
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utkarshakash said:

Homework Statement


If n is a positive integer, prove that [itex]2^n > 1+n\sqrt{2^{n-1}}[/itex]

Homework Equations



The Attempt at a Solution


I am thinking of applying AM GM HM inequality. But which numbers should I take to arrive at this inequality?

...This may not be correct for any number n that is a positive integer.

For example, [itex]2^{1}[/itex] is not greater than [itex]1 + n\sqrt{2^{1-1}}[/itex]. In fact, they are equivalent.
 
Mandelbroth said:
...This may not be correct for any number n that is a positive integer.

For example, [itex]2^{1}[/itex] is not greater than [itex]1 + n\sqrt{2^{1-1}}[/itex]. In fact, they are equivalent.

Ok so assume that it is 'greater that or equal to' instead of just 'is greater than' and prove it
 
utkarshakash said:
Ok so assume that it is 'greater that or equal to' instead of just 'is greater than' and prove it

No: YOU prove it, or at least show some effort towards the solution. Read the Forum rules!

RGV
 
Ray Vickson said:
No: YOU prove it, or at least show some effort towards the solution. Read the Forum rules!

RGV

Hey I really don't know how to solve this. I need some hints to get started. I've already stated that I am thinking of solving it using AM GM HM inequality. I know nothing more than this.
 
utkarshakash said:
Hey I really don't know how to solve this. I need some hints to get started. I've already stated that I am thinking of solving it using AM GM HM inequality. I know nothing more than this.

Use mathematical Induction. Do you know what is it ?
 
I think appying the binomial theorem to (1+ 1)n would also work.
 
sankalpmittal said:
Use mathematical Induction. Do you know what is it ?

Yes I know but I'm not required to use it. Also I'm not good at it.
 
utkarshakash said:
Yes I know but I'm not required to use it. Also I'm not good at it.

This might be your chance to improve on your technique then. Problems like this, beginning with "Prove for any n...", practically yell out "Use induction on me!" And mostly it's the easiest way to solve them.

I can see the binomial theorem being used to start 2n = (1+1)n = 1 + n Ʃ..., but getting from that sum to √2n-1 may be more work that induction would be. Maybe I'm overlooking some obvious trick though.
 
  • #10
utkarshakash said:
Yes I know but I'm not required to use it. Also I'm not good at it.

I understand what you want... But believe instead of using "Arithematic mean ≥ Geometric mean" , its more easier to use mathematical induction anyways...

But you want to use progression and series only... Ok , well , I throw you off a hint :

2n =2x2n-1.Now use A.M≥G.M here..
 

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