If x and y are positive

[Pathik]

If x and y are positive, which of the following must be greater than 1/root(x+y)

I. root(x+y)/2x

II [root(x) + root(y)] / (x+y)

III [ root(x) - root(y) ] / (x+y)

A. none
B. I only
C. II only
D. I and III
E II and III

What is the fastest way to solve this.

[Captain]

What is the fastest way to solve this.

I dont know a fastest way. But this is how I thought -
First I note that 1/root(x+y) is symmetric wrt x,y

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I. root(x+y)/2x

This is not symmetric with respect to x,y. So this cant be correct. What if I take a very small value of y; 1/root(x+y) goes to 1/root(x). but this equation goes to 1/(2root(x)). if x = 4 then 1/root(x+y) = 1/2, but I. becomes 1/4. Thus not always true.

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III [ root(x) - root(y) ] / (x+y)

This one is symmetric. so I will use x = y; III. becomes 0, and 1/root(x+y) becomes 1/root(2x); III will be less atleast in one case.

Now

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II [root(x) + root(y)] / (x+y)

This one is also symmetric. And putting x=y we will get II = 1/root(x) and the other as 1/ {root(2)*root(x)}, Thus II is greater.
At this point I declared II as the answer. To be sure I wrote both the expressions and proved II>1/root(x+y), for generic cases.

[RonPurewal]

http://www.manhattangmat.com/forums/post4838.html

[Guest]

Ron ,

I still feel lost with this Q..Stacey's explanation on the link you provided helped, but I don' get how to solve the problem. Please explain..

[RonPurewal]

Ron ,

I still feel lost with this Q..Stacey's explanation on the link you provided helped, but I don' get how to solve the problem. Please explain..

try this.