If X and Y are positive, which of the following must be greater than 1 / (x+y)^.5?

1. (X+Y)^.5 / 2x

2. [(X^.5) + (Y^.5)] / (x+y)

3. [(X^.5) - (Y^.5)] / (x+y)


Is the best way to do this problem by plugging in numbers? I noticed I was spending alot of time trying various values. Any recommendations?

Thanks!

The correct answer is 2.

Eq1 is not greater if Y=X .
Eq3 is always less.

THE FASTEST WAY IS TO EXPRESS EACH EQUATION AS A FUNCTION OF 1/(X+Y)^0.5

Please post the entire text of the question including answer choices. I'm assuming this is a Roman Numeral question?

We're told x and y are positive but not whether they are greater than 1, so I have to consider fractional possibilities. How do I know what to try?

When I take a square root:
Anything greater than 1 will get smaller (but remain larger than 1)
1 will stay the same
Anything between 0 and 1 will get bigger (but remain a fraction between 0 and 1)

When I take a reciprocal in each of the above cases:
1/something larger than 1 = something smaller than 1 (but still positive)
1/1 = 1
1/something smaller than 1 = something larger than 1

If I want to try numbers now, then I know I need to try a number from each set. Or I can continue with logic and the algebraic representations. Do whichever you are most comfortable with.

For trying numbers, first try something greater than 1:
x=2, y=2 (I'm trying the same numbers b/c I'm trying to see if I can prove things false and funny things happen when you use the same number for different variables). 1/(4)^.5 = 1/2.
Roman Numeral 1 (RN1): (4)^.5 / 2(2) = 2/4 = 1/2. Same, not greater, so elim RN1.
RN2: (2^.5 + 2^.5) / (4) = 2(2^.5) / 4. Well, 2^.5 is about 1.7. 2*1.7 = 3.4 / 4 = more than 1/2. So RN2 is okay, at least with this instance.
RN3: (2^.5 - 2^.5) / 4 = 0/4 = 0. Elim RN3.

At this point, I don't know whether I have to try more numbers b/c the answer choices haven't been listed. If I have both "none" and "II only" as options, then I have to try more numbers. If "none" is not an option, then I'm done.

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

I agree the entire question should be posted if there exists other equations to consider.

You will notice that equation 2 will always be more .
THE FASTEST WAY IS TO EXPRESS EACH EQUATION AS A FUNCTION OF 1/(X+Y)^0.5 . This aproach takes less than a minute.

The technical constraints of explaining this aspect on this forum is regretted.

a easy one(however, trying to eliminate the option by testing lots of numbers is hard to work out the question, wasting more time):
#1 for II, [(X^.5) + (Y^.5)] / (x+y) = {[(X^.5) + (Y^.5)] / (x+y)^.5} * [1 / (x+y)^.5]

#2 what we need to do is to kown wether the red part >1.

#3 square the red part, [x+y+2(xy)^.5]/(x+y) =1+[2(xy)^.5/(x+y)], so the red part>1.
(for III , the comparent red part is[x+y-2(xy)^.5]/(x+y) =1-[2(xy)^.5/(x+y)]<1)

so II is ok( III is not ok) !
done