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Author Topic:   Math proof... any idea?
teen4christ
Member (Idle past 5799 days)
Posts: 238
Joined: 01-15-2008


Message 1 of 10 (463341)
04-15-2008 2:06 PM


Someone asked me for help on proving this apparently simple algebraic problem. After doodling with it for several days, I still can't figure out a way to prove it. I'm sure it's simple, I just can't wrap my fingers around it. Any idea?
Prove that ((a+b)^2)/(4*a*b) > 1
I've plugged in numbers and numbers and they all came up to be greater than 1. But how do you go about proving this? I'm going crazy over this.

Replies to this message:
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kjsimons
Member
Posts: 821
From: Orlando,FL
Joined: 06-17-2003
Member Rating: 6.7


Message 2 of 10 (463347)
04-15-2008 2:25 PM
Reply to: Message 1 by teen4christ
04-15-2008 2:06 PM


Nevermind ... I screwed up!
Edited by kjsimons, : Brain wasn't functioning!

This message is a reply to:
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PaulK
Member
Posts: 17822
Joined: 01-10-2003
Member Rating: 2.2


Message 3 of 10 (463349)
04-15-2008 2:34 PM
Reply to: Message 1 by teen4christ
04-15-2008 2:06 PM


That should be >= 1 ! (It's 1 when a = b). And I think I've got it. How far have you got ? The early steps are pretty obvious.

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dwise1
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Posts: 5930
Joined: 05-02-2006
Member Rating: 5.8


Message 4 of 10 (463350)
04-15-2008 2:37 PM
Reply to: Message 1 by teen4christ
04-15-2008 2:06 PM


Are there restrictions on the range of values for a and b? For example, can one be negative and the other positive?
Assuming that they are always the same sign, then for
((a+b)^2)/(4*a*b) to be greater than one, it would need to be true that:
(a+b)^2 > 4*a*b
Now here's where my question about restrictions on the values of a and b comes in. (a+b)^2 is always positive, so if 4ab is negative (ie, if a and b have different signs), then
(a+b)^2 > 4*a*b
is true (a positive value is always greater than a negative).
However, if 4ab is negative, then
((a+b)^2)/(4*a*b)
is also negative, in which case it could not be greater than one.
Are there any restrictions on the values of a and b?
-------------------
PS
Take Two:
If ((a+b)^2)/(4*a*b) > 1, then (a+b)^2 > 4*a*b
a^2 + 2ab + b^2 > 4ab
a^2 - 2ab +b^2 > 0
(a-b)^2 > 0
a - b > 0
a > b
So we also have a requirement that a be greater than b?
Edited by dwise1, : add attempt at proof

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teen4christ
Member (Idle past 5799 days)
Posts: 238
Joined: 01-15-2008


Message 5 of 10 (463351)
04-15-2008 2:38 PM
Reply to: Message 2 by kjsimons
04-15-2008 2:25 PM


Sorry, I left out something. Both a and b are positive reals.

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NosyNed
Member
Posts: 8996
From: Canada
Joined: 04-04-2003


Message 6 of 10 (463354)
04-15-2008 2:50 PM
Reply to: Message 1 by teen4christ
04-15-2008 2:06 PM


Left out assumption
First I think you've left out an assumption:
a and b are not both equal to each other
If a and b = 1 then ((a+b)^2)/(4*a*b) = 1 (not > )
(a+b)^2 = a^2 + 2ab + b^2
4ab = 2(2ab)
subtract 2ab from both numerator and denominator
a^2 + b^2 ~ 2ab
If a = b then a^2 = b^2
so the numerator becomes
2a^2 and the denominator becomes 2aa = 2a^2 so the quotient is 1.

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dwise1
Member
Posts: 5930
Joined: 05-02-2006
Member Rating: 5.8


Message 7 of 10 (463355)
04-15-2008 2:51 PM
Reply to: Message 5 by teen4christ
04-15-2008 2:38 PM


Do a and b also need to be not equal?
if b = a, then
((a+b)^2)/(4*a*b)
is
((a+a)^2)/(4*a*a)
((2a)^2)/(4*a^2)
(4a^2)/(4a^2)
4a^2 = 4a^2
therefore, (4a^2)/(4a^2) = 1
Therefore, ((a+b)^2)/(4*a*b) >1 is false.

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teen4christ
Member (Idle past 5799 days)
Posts: 238
Joined: 01-15-2008


Message 8 of 10 (463356)
04-15-2008 2:52 PM


I figured it out.
(a+b)^2 <= 4ab
a^2+2ab+b^2 <= 4ab
a^2-2ab+b^2 <= 0
(a-b)^2 <= 0 NOT TRUE!!!
Therefore, since (a-b)^2 > 0 always, (a+b)^2 > 4ab.

Replies to this message:
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NosyNed
Member
Posts: 8996
From: Canada
Joined: 04-04-2003


Message 9 of 10 (463358)
04-15-2008 3:03 PM
Reply to: Message 1 by teen4christ
04-15-2008 2:06 PM


Not Equal then true
If a and b are not equal then one is greater than the other by a positive amount say d.
This is symmetrical so let's make a = b + d
now we have
(b + d)^2 + b^2 > 2(b+d)b
b^2 +2bd + d^2 + b^2 > 2b^2 + 2bd
and d^2 is > 0 as defined
so the left is > the right
It is true when not equal a and b

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Chiroptera
Inactive Member


Message 10 of 10 (463360)
04-15-2008 3:23 PM
Reply to: Message 8 by teen4christ
04-15-2008 2:52 PM


You can reverse your steps to get a direct proof.
(a - b)2 >= 0
a2 - 2ab + b2 >= 0
a2 - 2ab + b2 + 4ab >= 0 + 4ab
a2 + 2ab + b2 >= 4ab
(a + b)2 >= 4ab
(a + b)2/(4ab) >= 1
where, for the last line to be valid, 4ab must be positive, that is, a and b must have the same sign.

Speaking personally, I find few things more awesome than contemplating this vast and majestic process of evolution, the ebb and flow of successive biotas through geological time. Creationists and others who cannot for ideological or religious reasons accept the fact of evolution miss out a great deal, and are left with a claustrophobic little universe in which nothing happens and nothing changes.
-- M. Alan Kazlev

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