IF a,b >0

Then prove: AM -HM<= (b-a)^2/4a

This is what I did

AM = (a + b)/2

HM = 2/(1/a + 1/b)

So when I subtract AM - HM: I got this

(a+b)/2 - 2ab/(a+b)

[(a+b)^2 - 4ab]/[2(a+b)]

which reduces to [(a-b)^2] / [2(a+b)]

Now what do I do?

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- Jun 5th 2006, 06:06 PMNichelle14Need help on a proof
IF a,b >0

Then prove: AM -HM<= (b-a)^2/4a

This is what I did

AM = (a + b)/2

HM = 2/(1/a + 1/b)

So when I subtract AM - HM: I got this

(a+b)/2 - 2ab/(a+b)

[(a+b)^2 - 4ab]/[2(a+b)]

which reduces to [(a-b)^2] / [2(a+b)]

Now what do I do? - Jun 5th 2006, 06:21 PMmalaygoelQuote:

Originally Posted by**Nichelle14**

if a< b

2a< a+b

1/2a> 1/(a+b)

and your proof is complete - Jun 5th 2006, 06:32 PMNichelle14
I believe you are asking for this realationship a and b are positive.

I don't understand how your part of the proof makes it complete?

In other words, how does your part be <= (b-a)^2/4b :confused:

Can you explain some more? - Jun 5th 2006, 06:38 PMmalaygoelQuote:

Originally Posted by**Nichelle14**

continuing my work

1/2a > 1/(a+b)

multiplying both sides by [(b-a)^2]/2 (which is a positive quantity)

we get

(b-a)^2/4a >(b-a)^2/2(a+b) RHS is equal to AM-HM - Jun 6th 2006, 06:08 PMNichelle14
earlier you posted

if a < b, then 2a < a + b

Why did you chose to do this? - Jun 6th 2006, 06:43 PMmalaygoelQuote:

Originally Posted by**Nichelle14**

I had the choice whether a would be smaller or bigger than b because had not mentioned anything.

I chose a<b because it impies the equation you have to prove. - Jun 6th 2006, 06:43 PMThePerfectHacker
It is not true.

What about $\displaystyle a=b=2$ :eek:

-----

I believe I found a solution.

As my top sentence says, your inequality is false.

It is however true when, for $\displaystyle x,y>0$.

$\displaystyle \phi y\geq x$

Where,

$\displaystyle \phi $ is the solution to the cubic,

$\displaystyle 2x^3+x^2+3x-1=0$. - Jun 6th 2006, 07:23 PMmalaygoelQuote:

Originally Posted by**ThePerfectHacker**