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IanStewart
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If x and y are positive, which of the following must be greater than 16 Mar 2023, 13:55
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Dipanjan005
IanStewart please explain why the 3rd option is wrong?
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If x = y, say, then the third option will be equal to zero, so will always be less than 1/√(x + y), since that's positive. The third option can even be negative if y > x.
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If x and y are positive, which of the following must be greater than 08 Nov 2024, 05:00
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If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

I. \(\frac{\sqrt{x+y}}{2x}\)

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) I only
(C) II only
(D) I and III only
(E) II and III only

Show Spoiler
Attachment:
PS.JPG
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In comparison, we always need to look for something common. To compare fractions, either numerator or denominator must be the same.


I. \(\frac{\sqrt{x+y}}{2x}\) Let's compare it with our given expression.

\(\frac{1}{\sqrt{x+y}} = \frac{\sqrt{x+y}}{x+y}\) (Dividing and multiply by sqrt(x+y))

Their numerators are same so now we just compare denominators. Is 2x ( = x + x) smaller or greater than (x+y)? That simply depends on whether y is smaller or greater than x. Hence this expression can be greater or smaller than the given expression.

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) Let's compare it with our given expression.

\(\frac{1}{\sqrt{x+y}} = \frac{\sqrt{x+y}}{x+y}\) (Dividing and multiply by sqrt(x+y))

Here denominator is same so we compare numerators. If you recall that \(\sqrt{x+y} \leq \sqrt{x}+\sqrt{y}\) then great. Equality holds when x or y is 0. Since x and y are positive, this expression will always be greater.

Else try some values. It seems likely. But we still need to convince ourselves that it will work in all cases.
\((\sqrt{x}+\sqrt{y})^2 = x + y + 2\sqrt{xy}\)
\((\sqrt{x+y})^2 = x + y\)

So we see that \(\sqrt{x}+\sqrt{y}\) is greater.

Hence II is greater in all cases.


III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

Since there is a negative in the numerator, it is easy to eliminate this option. It can be negative. Say if x = 1 and y = 4. Our given expression cannot be negative.


Answer (C)
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If x and y are positive, which of the following must be greater than 11 Jan 2026, 21:10
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Looks like I am doing something wrong which I cant understand. I took x=6 and y =3. That way in the second statement I get root 3 + root 6= root 9/ root 9 which gives me 1/3 which is similar to the initial equation which is 1/3. Hence I get that statement must always not be greater. What am I doing wrong?
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If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

I. \(\frac{\sqrt{x+y}}{2x}\)

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) I only
(C) II only
(D) I and III only
(E) II and III only

Show Spoiler
Attachment:
PS.JPG

Solving this question helps. Taking a timed set of similar questions in GMAT Club Forum Quiz → is even better.
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If x and y are positive, which of the following must be greater than 11 Jan 2026, 22:39
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Looks like I am doing something wrong which I cant understand. I took x=6 and y =3. That way in the second statement I get root 3 + root 6= root 9/ root 9 which gives me 1/3 which is similar to the initial equation which is 1/3. Hence I get that statement must always not be greater. What am I doing wrong?

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If x = 6 and y = 3, then:

\(\frac{1}{\sqrt{x+y}}= \frac{1}{\sqrt{6+3}}=\frac{1}{3} \)

\(\frac{\sqrt{x}+\sqrt{y}}{x+y}= \frac{\sqrt{6}+\sqrt{3}}{6+3}= \frac{\sqrt{6}+\sqrt{3}}{9}\approx {0.46}\).

The point is \(\sqrt{6}+\sqrt{3}\neq\sqrt{9}\).

Hope it helps.
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If x and y are positive, which of the following must be greater than 14 Jan 2026, 01:19
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Thank you. Had to brush up my basics on what is allowed for roots :)
Bunuel


If x = 6 and y = 3, then:

\(\frac{1}{\sqrt{x+y}}= \frac{1}{\sqrt{6+3}}=\frac{1}{3} \)

\(\frac{\sqrt{x}+\sqrt{y}}{x+y}= \frac{\sqrt{6}+\sqrt{3}}{6+3}= \frac{\sqrt{6}+\sqrt{3}}{9}\approx {0.46}\).

The point is \(\sqrt{6}+\sqrt{3}\neq\sqrt{9}\).

Hope it helps.
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If x and y are positive, which of the following must be greater than 24 Jun 2026, 04:34
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one question Bunuel,
to rationalize the denominator, we multiply both the terms by its conjugate
here the conjugate of 1/sqrt(x+y) isn't sqrt(x-y)?


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If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

I. \(\frac{\sqrt{x+y}}{2x}\)

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

III. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) I only
(C) II only
(D) I and III only
(E) II and III only

Show Spoiler
Attachment:
PS.JPG
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If x and y are positive, which of the following must be greater than 24 Jun 2026, 07:26
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SwethaReddyL
one question Bunuel,
to rationalize the denominator, we multiply both the terms by its conjugate
here the conjugate of 1/sqrt(x+y) isn't sqrt(x-y)?



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No. A conjugate applies to a two-term expression, such as \(\sqrt{x}+\sqrt{y}\), whose conjugate is \(\sqrt{x}-\sqrt{y}\).

For \(\frac{1}{\sqrt{x+y}}\) the denominator is just one radical, \(\sqrt{x+y}\). To rationalize it, we multiply by \(\frac{\sqrt{x+y}}{\sqrt{x+y}}\), not by \(\sqrt{x-y}\). This gives:

\(\frac{1}{\sqrt{x+y}} * \frac{\sqrt{x+y}}{\sqrt{x+y}} = \frac{\sqrt{x+y}}{x+y}\)
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If x and y are positive, which of the following must be greater than 25 Jun 2026, 04:05
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got it, thank you so much.
Bunuel

No. A conjugate applies to a two-term expression, such as \(\sqrt{x}+\sqrt{y}\), whose conjugate is \(\sqrt{x}-\sqrt{y}\).

For \(\frac{1}{\sqrt{x+y}}\) the denominator is just one radical, \(\sqrt{x+y}\). To rationalize it, we multiply by \(\frac{\sqrt{x+y}}{\sqrt{x+y}}\), not by \(\sqrt{x-y}\). This gives:

\(\frac{1}{\sqrt{x+y}} * \frac{\sqrt{x+y}}{\sqrt{x+y}} = \frac{\sqrt{x+y}}{x+y}\)
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If x and y are positive, which of the following must be greater than 25 Jun 2026, 07:11
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Hi SwethaReddyL,

You've mixed up two different "rationalizing" situations, and once you separate them the step in Bunuel's solution will click.

Conjugates are only for two-term (binomial) denominators. When the denominator looks like a + √b, you multiply by a − √b so the cross terms cancel through difference of squares:

- (a + √b)(a − √b) = a2 − b - radical gone.

Here the denominator is just √(x+y). That is a single square root - one term, not two. There is nothing to "pair off," so no conjugate is needed at all.

To clear a lone square root, just multiply by that same root:

- 1/√(x+y) × √(x+y)/√(x+y) = √(x+y)/(x+y)

Because √(x+y) · √(x+y) = x+y, the radical leaves the denominator. That's exactly the move in the solution - multiply top and bottom by √(x+y), nothing fancier.

Why √(x−y) is the wrong instinct: a conjugate flips the sign between two terms. But x+y sits together under one root - it is not √x + √y, so there's no + to flip into a −. (In fact √(x+y) ≠ √x + √y, which is the very same trap Varun hit higher up in the thread.)

Answer: C

SwethaReddyL
one question Bunuel,
to rationalize the denominator, we multiply both the terms by its conjugate
here the conjugate of 1/sqrt(x+y) isn't sqrt(x-y)?



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