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

[utkarshakash]

Homework Statement

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

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?

 

[Mandelbroth]

Homework Statement

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

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, 2^{1} is not greater than 1 + n\sqrt{2^{1-1}}. In fact, they are equivalent.

 

[utkarshakash]

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

For example, 2^{1} is not greater than 1 + n\sqrt{2^{1-1}}. 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

 

[Ray Vickson]

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

 

[utkarshakash]

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.

 

[sankalpmittal]

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 ?

 

[HallsofIvy]

I think appying the binomial theorem to (1+ 1)ⁿ would also work.

 

[utkarshakash]

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.

 

[Michael Redei]

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 √2^n-1 may be more work that induction would be. Maybe I'm overlooking some obvious trick though.

 

[sankalpmittal]

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 =2x2^n-1.Now use A.M≥G.M here..