# Need help on a proof

• Jun 5th 2006, 06:06 PM
Nichelle14
Need 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 PM
malaygoel
Quote:

Originally Posted by Nichelle14
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?

Is there any relation between a and b?
if a< b
2a< a+b
1/2a> 1/(a+b)
• Jun 5th 2006, 06:32 PM
Nichelle14
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 PM
malaygoel
Quote:

Originally Posted by Nichelle14
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

Can you explain some more?

originally you have posted (b-a)^2/4a
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 PM
Nichelle14
earlier you posted

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

Why did you chose to do this?
• Jun 6th 2006, 06:43 PM
malaygoel
Quote:

Originally Posted by Nichelle14
earlier you posted

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

Why did you chose to do this?

a and b are the two consecutive terms of a series and they should not be equal
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 PM
ThePerfectHacker
It is not true.
What about $a=b=2$ :eek:
-----

I believe I found a solution.
As my top sentence says, your inequality is false.

It is however true when, for $x,y>0$.
$\phi y\geq x$
Where,
$\phi$ is the solution to the cubic,
$2x^3+x^2+3x-1=0$.
• Jun 6th 2006, 07:23 PM
malaygoel
Quote:

Originally Posted by ThePerfectHacker
It is not true.
What about $a=b=2$ :eek:
-----

I believe I found a solution.
As my top sentence says, your inequality is false.

It is however true when, for $x,y>0$.
$\phi y\geq x$
Where,
$\phi$ is the solution to the cubic,
$2x^3+x^2+3x-1=0$.

I don,t think my inequality is false. If you see what we set to prove it includes >= sign. It can be broken into > and = separately. If a=b= any positive real number, AM=GM=HM=a=b which is not of much importance. The real thing lies in inequality part. Hence we have to use a relation between the two terms. However, I don't understand your solution, please explain it.