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If a and b are positive, is (a^-1 + b^-1)^-1 less than

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If a and b are positive, is (a^-1 + b^-1)^-1 less than [#permalink] New post 28 May 2006, 22:49
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If a and b are positive, is (a^-1 + b^-1)^-1 less than (a^-1*b^-1)^-1?

(1) a = 2b
(2) a + b > 1
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 [#permalink] New post 28 May 2006, 23:24
Both the statements are sufficient in themselves.

The equation becomes :

{ (a * b) / (a + b ) } < (a * b)

Thus if :
a) a = 2b,
then eqn is ==> ( (2/3) * b) < (2 * b^2) which is true.

b) a + b=1, then LHS and RHS become equal.

Last edited by buzzgaurav on 29 May 2006, 00:11, edited 2 times in total.
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 [#permalink] New post 29 May 2006, 00:12
shobhitb wrote:
buzzgaurav wrote:
Both the statements are sufficient in themselves.

The equation becomes :

{ (a * b) / (a + b ) } < (a * b)

Thus if :
a) a = 2b,
then eqn is ==> b < (2 * b^2) which is true.

b) a + b=1, then also LHS is smaller than RHS due to a factor (a + b) in the denominator of LHS.


I am getting B

Sorry for the mistake in the original post, but I am still getting B.
I might be wrong as I am not at all good at quant :?
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 [#permalink] New post 29 May 2006, 05:21
Is ab/(a+b) < ab? given a and b are positive...
This is only possible if Denominator a+b > 1

2. SUFF.
1. If a+b < 1, say a = 0.2 and b = 0.1 , answer is No
if a+b > 1, say a = 2 , b=1, answer is Yes.
INSUFF.

B iti is!
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 [#permalink] New post 29 May 2006, 07:12
Yeah. B is the answer. Try plugging in 0.1 and 5 for a where a=2b and we can eliminate the answer.
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 [#permalink] New post 29 May 2006, 10:30
OA is indeed B.

Great explanation Giddi
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 [#permalink] New post 29 May 2006, 18:19
"B" it is.

Girish, I really appreciate your approach to problems :)
  [#permalink] 29 May 2006, 18:19
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