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# Stumped on this one ..

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Manager
Joined: 29 Nov 2006
Posts: 56
Followers: 1

Kudos [?]: 0 [0], given: 0

Stumped on this one .. [#permalink]  23 Jun 2007, 13:38
Stumped on this one ...
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Senior Manager
Joined: 28 Feb 2007
Posts: 306
Followers: 1

Kudos [?]: 32 [0], given: 0

1/(Sqrt (X+Y))=Sqrt(X+Y)/(X+Y)

[Sqrt(X)+Sqrt(Y)]/(X+Y) >= Sqrt(X+Y)/(X+Y)

First as the denominator is same u can drop.
Everything depends on the Numerator.to check just take square of both sides (numerator is enough):

X+Y+2*Sqrt(x*y)>= X+Y
then 2*Sqrt(x*y)>=0 because X, Y are Positive.
Senior Manager
Joined: 27 Jul 2006
Posts: 298
Followers: 1

Kudos [?]: 7 [0], given: 0

I am going with 1 and 2 only.

3 is actually equal, and so cannot be greater.
Senior Manager
Joined: 04 Mar 2007
Posts: 442
Followers: 1

Kudos [?]: 16 [0], given: 0

Ref (I)
let x=1 and y = 1
1/sqrt(2)=sqrt(2)/2
Hence it's not (I)
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