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# If x and y are positive, which of the following must be

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Director
Joined: 22 Mar 2011
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23 Oct 2012, 10:31
1
1
voodoochild wrote:
Experts,
I liked Walker's method. I chose X = y =1 and got 'none'. However, can we solve this question algebraically?

[Note: x and y are +ve; hence both the sides of inequalities are positive. There shouldn't be any harm in squaring]

#1 - This is straightforward.
For #2 - here's what I did: (Sqrt (x) + Sqrt (y))/2 > 1/(Sqrt(x+y))
Square both sides:
x + y + 2*sqrt(xy) > 24/(x+y)
OR

(x+y)^2 + 2(x+y)Sqrt (xy) > 4

This seems to hold good only when (x+y)^2 is greater than 2. (I am not sure about this one)

III:-

Simplifying, sqrt(x)-sqrt(y)>sqrt(x+y)
Now, x and y are positive. Therefore, rHS >0 => Sqrt (x) > sqrt(y)

Also, squaring both the sides, x+y-2sqrt(xy) > x+y => -2sqrt(xy)>0 ====> Impossible for any value of x and y because x and y are positive.

Is this method correct? Please let me know.

[Disclaimer: I wouldn't use this method on the GMAT. This is just for learning.]

II: (x+y)^2 + 2(x+y)Sqrt (xy) > 4
This seems to hold good only when (x+y)^2 is greater than 2.

Why try to figure out when this inequality holds? It is easier to see when it doesn't hold. For example for x and y very close to 0.

III: You cannot square the inequality because you don't know whether x is greater than y or not.
If x < y, the left hand side is negative, and obviously it cannot be greater than a positive number on the right hand side.
You got "lucky" here, the conclusion is right, but the way you did it, is wrong. Never ever square an expression if you are not absolutely sure that it is positive.

Walker made a good choice, x = y - clearly the inequality cannot hold.
Instead of squaring, you can also try another way:
Since x and y are both positive, $$\sqrt{x+y}>\sqrt{x}$$, therefore, from $$\sqrt{x}-\sqrt{y}>\sqrt{x+y}$$ we would get that $$\sqrt{x}-\sqrt{y}>\sqrt{x}$$ or $$-\sqrt{y}>0$$, impossible.
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Re: If x and y are positive, which of the following must be  [#permalink]

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23 Oct 2012, 10:56
EvaJager wrote:
Why try to figure out when this inequality holds? It is easier to see when it doesn't hold. For example for x and y very close to 0.

III: You cannot square the inequality because you don't know whether x is greater than y or not.
If x < y, the left hand side is negative, and obviously it cannot be greater than a positive number on the right hand side.
You got "lucky" here, the conclusion is right, but the way you did it, is wrong. Never ever square an expression if you are not absolutely sure that it is positive.

I don't agree because in sqrt(x)-sqrt(y)>sqrt(x+y), x+y is positive. We know that sqrt (x+y) is positive. HEnce, LHS has to be positive. It can't be -ve. X cannot be < Y. That's why both the sides of the inequalities are positive. Hence, we should be able to square them.

Thoughts?
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Re: If x and y are positive, which of the following must be  [#permalink]

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Updated on: 24 Oct 2012, 01:17
1
voodoochild wrote:
EvaJager wrote:
Why try to figure out when this inequality holds? It is easier to see when it doesn't hold. For example for x and y very close to 0.

III: You cannot square the inequality because you don't know whether x is greater than y or not.
If x < y, the left hand side is negative, and obviously it cannot be greater than a positive number on the right hand side.
You got "lucky" here, the conclusion is right, but the way you did it, is wrong. Never ever square an expression if you are not absolutely sure that it is positive.

I don't agree because in sqrt(x)-sqrt(y)>sqrt(x+y), x+y is positive. We know that sqrt (x+y) is positive. HEnce, LHS has to be positive. It can't be -ve. X cannot be < Y. That's why both the sides of the inequalities are positive. Hence, we should be able to square them.

Thoughts?

If your chain of thoughts was "if ... (inequality holds) ... then (something must be positive AND I should keep in mind this condition)...then (I can square)...then (I reach something which is not true)...therefore (my initial assumption is wrong, meaning the inequality doesn't necessarily hold)" - in fact is OK. The condition is that x must be greater than y. If the chain of reasoning would have reached at some point a conclusion stating that x is less than y, would you have been able to reach a conclusion, I mean, not to forget, that you started with the opposite?

My point is that if you see on a question a difference, $$\sqrt{x}-\sqrt{y}$$ which should be positive, why would you continue to do extra work, like squaring? If x is not necessarily greater than y (it is not given), then you should stop. Definitely, the purpose of this question was not to test your algebraic abilities of squaring expressions. You shouldn't do this even when just practicing and learning. The most important thing you can learn from this exercise, is that there are valuable clues on the way and you should develop skills not to miss them. If you just continue like a robot, "there are square roots, we should get rid of them, we should therefore square expressions..." , you don't help your preparation for the real test at all. And with statement II, you could clearly see that squaring and algebraic manipulations don't help, they can even complicate the expressions and make it more difficult to reach a conclusion.
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Originally posted by EvaJager on 23 Oct 2012, 12:19.
Last edited by EvaJager on 24 Oct 2012, 01:17, edited 1 time in total.
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Re: If x and y are positive, which of the following must be  [#permalink]

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23 Oct 2012, 19:24
1
voodoochild wrote:
For #2 - here's what I did: (Sqrt (x) + Sqrt (y))/2 > 1/(Sqrt(x+y))
Square both sides:
x + y + 2*sqrt(xy) > 2/(x+y)
OR

(x+y)^2 + 2(x+y)Sqrt (xy) > 4

This seems to hold good only when (x+y)^2 is greater than 2. (I am not sure about this one)

A point to note: From where did this inequality come?
$$( \sqrt{x} + \sqrt{y})/2 > 1/\sqrt{(x+y)}$$
It isn't given to you. They are asking you whether this inequality holds in case x and y are positive.

So what you are trying to figure out is this:
Is $$( \sqrt{x} + \sqrt{y})/2 > 1/\sqrt{(x+y)}$$?
Since x and y are both positive, both sides of the inequality are positive (on their own). You can square it if you really wish to. It's something like 'is a > b?' given both a and b are positive. If a^2 is greater than b^2, then obviously, a is greater than b.

voodoochild wrote:
III:-

Simplifying, sqrt(x)-sqrt(y)>sqrt(x+y)
Now, x and y are positive. Therefore, rHS >0 => Sqrt (x) > sqrt(y)

Also, squaring both the sides, x+y-2sqrt(xy) > x+y => -2sqrt(xy)>0 ====> Impossible for any value of x and y because x and y are positive.

Is $$\sqrt{x} - \sqrt{y}> \sqrt{(x+y)}$$?

Mind you, it is not given to you that the inequality holds. You need to find out whether it does. The LHS can be negative here (in case root x is less than root y) on its own. In that case, the inequality certainly doesn't hold.

It is something like this:
Is a > b?
b is positive. You don't know about a. You can't square the inequality. You cannot assume that a must be positive since b is positive.
Say, if a = -4 and b = 2
(-4)^2 > 2^2 holds even though a is not greater than b. So even if you get that a^2 is in fact greater than b^2, you cannot deduce that a must be greater than b.

Squaring in such a case will only lead to a ton of confusion. Avoid it.
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Re: If x and y are positive, which of the following must be  [#permalink]

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24 Oct 2012, 10:00
VeritasPrepKarishma wrote:
Is $$\sqrt{x} - \sqrt{y}> \sqrt{(x+y)}$$?

Mind you, it is not given to you that the inequality holds. You need to find out whether it does. The LHS can be negative here (in case root x is less than root y) on its own. In that case, the inequality certainly doesn't hold.

Karishma,

#1- We are assuming that the given inequality is TRUE, and then evaluating its correctness. Hence, if x and y are positive, sqrt(x)-sqrt(y) will be positive. If it's -ve then the inequality cannot hold good. That's what happened in the end.
#2 - One of the best strategies on MUST BE TRUE questions is to solve the equation as much as possible. There are questions on which plugging-in numbers won't work at all.

Thoughts?
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Re: If x and y are positive, which of the following must be  [#permalink]

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24 Oct 2012, 12:23
1
voodoochild wrote:
VeritasPrepKarishma wrote:
Is $$\sqrt{x} - \sqrt{y}> \sqrt{(x+y)}$$?

Mind you, it is not given to you that the inequality holds. You need to find out whether it does. The LHS can be negative here (in case root x is less than root y) on its own. In that case, the inequality certainly doesn't hold.

Karishma,

#1- We are assuming that the given inequality is TRUE, and then evaluating its correctness. Hence, if x and y are positive, sqrt(x)-sqrt(y) will be positive. If it's -ve then the inequality cannot hold good. That's what happened in the end.
#2 - One of the best strategies on MUST BE TRUE questions is to solve the equation as much as possible. There are questions on which plugging-in numbers won't work at all.

Thoughts?

#1 There is a question: which must be greater...None of the expressions is guaranteed to be greater than the given expression.
$$\sqrt{x}-\sqrt{y}$$ is positive if and only if $$x>y$$. Is this guaranteed? NO! No restriction on $$x$$ and $$y$$, either one can be greater than the other one, or they can also be equal. Even without further checking, no matter what is the conclusion from further analysis, if the initial condition of $$x>y$$ is not guaranteed to hold, the inequality must not hold!!! You got "lucky", because finally it turned out that the inequality cannot hold. But, just hypothetically, if you would have found that the inequality can hold if $$x>y$$? What would have been your answer? That the inequality MUST hold? Again a big NO!!! Because the initial condition of $$x>y$$ is not guaranteed to hold!!! In conclusion, it's not worth the whole trouble to check all kinds of other conditions when already the first one must not hold.

#2 Big NO!!! How did you reach this conclusion?
Many times in mathematics, to a question whether a certain inequality holds for every value of the variable, it is much easier to find even one particular value for which it doesn't hold, to state with certainty that the inequality doesn't hold for all the values of the variable. As it was the case here in this question. The expressions were deliberately given complicated, with square roots. The test was on whether you are able to evaluate the values of different expressions, to "feel" when one is the smallest, when another one is the greatest...
GMAT is testing the flexibility of your mind. You don't do any good to yourself locking your mind on inefficient methods.
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Re: If x and y are positive, which of the following must be  [#permalink]

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24 Oct 2012, 17:13
EvaJager wrote:
#1 There is a question: which must be greater...None of the expressions is guaranteed to be greater than the given expression.
$$\sqrt{x}-\sqrt{y}$$ is positive if and only if $$x>y$$. Is this guaranteed? NO! No restriction on $$x$$ and $$y$$, either one can be greater than the other one, or they can also be equal. Even without further checking, no matter what is the conclusion from further analysis, if the initial condition of $$x>y$$ is not guaranteed to hold, the inequality must not hold!!! You got "lucky", because finally it turned out that the inequality cannot hold. But, just hypothetically, if you would have found that the inequality can hold if $$x>y$$? What would have been your answer? That the inequality MUST hold? Again a big NO!!! Because the initial condition of $$x>y$$ is not guaranteed to hold!!! In conclusion, it's not worth the whole trouble to check all kinds of other conditions when already the first one must not hold.

Eva,
I still disagree. Either I am missing something very crucial, or I am not able to express myself properly. I would lean towards the former until the latter is ruled out. Why are we ignoring the RHS side of the inequality? sqrt(x) - sqrt(y) as it is, could be >, =, < 0. I agree. However, with sqrt (x-y) on the RHS, that expression must be greater than zero. if a> sqrt (x) ===> a has to be greater than or equal to 0, irrespective of the value of x. (Let's assume that x is not a complex number ) In the next step, I would do the analysis that you have described : whether X < Y, or X<1 and Y<1 etc. However, while squaring, I can safely assume that sqrt(x) > sqrt(y) because the RHS is +ve.

the main objective in this question is to evaluate whether the inequality MUST BE TRUE. The inequality as it is, could be squared. I am not able to think of an inequality in which I will be 'unlucky'

Regarding #2, we cannot solve many MUST BE TRUE questions using number plugging. Example: if-x-x-x-which-of-the-following-must-be-true-about-x-68886.html In this one, substitution will not work. (try x=0.5 ). One has to solve inequalities to find the roots.

I could be wrong.

Thoughts?

thanks
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Re: If x and y are positive, which of the following must be  [#permalink]

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24 Oct 2012, 21:21
1
voodoochild wrote:
VeritasPrepKarishma wrote:
Is $$\sqrt{x} - \sqrt{y}> \sqrt{(x+y)}$$?

Mind you, it is not given to you that the inequality holds. You need to find out whether it does. The LHS can be negative here (in case root x is less than root y) on its own. In that case, the inequality certainly doesn't hold.

Karishma,

#1- We are assuming that the given inequality is TRUE, and then evaluating its correctness. Hence, if x and y are positive, sqrt(x)-sqrt(y) will be positive. If it's -ve then the inequality cannot hold good. That's what happened in the end.
#2 - One of the best strategies on MUST BE TRUE questions is to solve the equation as much as possible. There are questions on which plugging-in numbers won't work at all.

Thoughts?

Having interacted with you over the past few months, I think that you genuinely care about having strong concepts. Hence, I will try to explain in detail the points you are missing (in my opinion):

#2 - One of the best strategies on MUST BE TRUE questions is to solve the equation as much as possible. There are questions on which plugging-in numbers won't work at all.

Absolutely! I dislike number plugging. For most questions, I follow the strategy of thinking logically; I try to apply conceptual understanding to questions. That said, in some questions, number plugging is what works. For 'MUST BE TRUE' questions, number plugging is useless if the relation actually must be true. You cannot plug in every number and check to establish that the relation is indeed true. So once you get that the relation holds for numbers from different ranges, you need to start thinking logically why it must hold for all numbers. On the other hand, if you can find one number for which the relation doesn't hold true, you are done! You already know that the relation doesn't hold for all numbers. Hence, in some cases, the most successful strategy is when you can see that the relation does not hold for an easy number e.g. 0 or 1 etc. Hence, number plugging has its uses, limited they may be.
In the context of this question, what I see is that you are removing the roots and then plugging in numbers (in statement II). Removing the roots didn't actually give you the range, right? Statement III worked out well but it was a fluke. I will explain this more in your next point.

#1- We are assuming that the given inequality is TRUE, and then evaluating its correctness. Hence, if x and y are positive, sqrt(x)-sqrt(y) will be positive. If it's -ve then the inequality cannot hold good. That's what happened in the end.

Fine. Let's assume that the inequality holds (statement III). We get that LHS must be positive. We do some algebra and we get that -2sqrt(xy)>0 which we know is not possible. This proves to us that the inequality definitely does not hold i.e. LHS is not greater than RHS for any value of x and y (x and y are both positive). The reason why this worked out is that the inequality does not hold for any value of x and y i.e. it is definitely false. It is possible that you get an inequality which holds for some values and does not for others. In that case, you would again need to plug in numbers and check (like you did in statement 2). Our question was whether the inequality holds for all values. In statement 3, by assuming that it does, we were able to prove that it doesn't hold for any value. The strategy fails in case it holds for some values and does not for others (like it did for statement 2).

Now, what irks me about assuming that the inequality holds and squaring? Let me show you using a simple example.
Say, my question is: Is A > B?
(A and B are expressions using x and y. Say, the values A can take are A>2 or A < -2 and the values B can take are 0 < B < 2. But of course, this is not given to you and the expressions are complex so you cant really see it either)

You assume that A > B and square it.
A^2 > B^2
Not you plug in values for x and y in the inequality and you see that for every value you put in, A^2 is greater than B^2 (it will be because of the range of values A and B can take). What does it imply? Does it mean that A > B for every value of x and y? But it is not true. A is not greater than B for every value of x and y. Hence, the entire exercise could be misleading.

Bottom line: Try to avoid assuming that the relation questioned actually holds true. (It might work splendidly in rare situations but more often than not, it will confuse you to no end)
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Re: If x and y are positive, which of the following must be  [#permalink]

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25 Oct 2012, 01:10
1
voodoochild wrote:
EvaJager wrote:
#1 There is a question: which must be greater...None of the expressions is guaranteed to be greater than the given expression.
$$\sqrt{x}-\sqrt{y}$$ is positive if and only if $$x>y$$. Is this guaranteed? NO! No restriction on $$x$$ and $$y$$, either one can be greater than the other one, or they can also be equal. Even without further checking, no matter what is the conclusion from further analysis, if the initial condition of $$x>y$$ is not guaranteed to hold, the inequality must not hold!!! You got "lucky", because finally it turned out that the inequality cannot hold. But, just hypothetically, if you would have found that the inequality can hold if $$x>y$$? What would have been your answer? That the inequality MUST hold? Again a big NO!!! Because the initial condition of $$x>y$$ is not guaranteed to hold!!! In conclusion, it's not worth the whole trouble to check all kinds of other conditions when already the first one must not hold.

Eva,
I still disagree. Either I am missing something very crucial, or I am not able to express myself properly. I would lean towards the former until the latter is ruled out. Why are we ignoring the RHS side of the inequality? sqrt(x) - sqrt(y) as it is, could be >, =, < 0. I agree. However, with sqrt (x-y) on the RHS, that expression must be greater than zero. if a> sqrt (x) ===> a has to be greater than or equal to 0, irrespective of the value of x. (Let's assume that x is not a complex number ) In the next step, I would do the analysis that you have described : whether X < Y, or X<1 and Y<1 etc. However, while squaring, I can safely assume that sqrt(x) > sqrt(y) because the RHS is +ve.

the main objective in this question is to evaluate whether the inequality MUST BE TRUE. The inequality as it is, could be squared. I am not able to think of an inequality in which I will be 'unlucky'

Regarding #2, we cannot solve many MUST BE TRUE questions using number plugging. Example: if-x-x-x-which-of-the-following-must-be-true-about-x-68886.html In this one, substitution will not work. (try x=0.5 ). One has to solve inequalities to find the roots.

I could be wrong.

Thoughts?

thanks

You wrote:
"the main objective in this question is to evaluate whether the inequality MUST BE TRUE."

Meaning of the word whether:

whether
Conjunction:
Expressing a doubt or choice between alternatives: "he seemed undecided whether to go or stay".
Expressing an inquiry or investigation (often used in indirect questions): "I'll see whether she's at home".

Look at the following question:
"If $$a,\,b$$ and $$c$$ are non-zero integers, must $$a - b$$ be greater than $$c^2$$?" What would be your approach?
You don't need to square here anything. So, what's your reasoning? Or what should be your reasoning?
You could say that $$c^2$$ is always positive, then $$a-b$$ must be positive, but this can be true or not.
Therefore, the answer should be NO, $$a-b$$ must not necessarily be greater than $$c^2$$.

So why cannot you apply the above reasoning when instead of $$a-b$$ you have $$\sqrt{x}-\sqrt{y}$$?
The first stage is the same, you conclude that $$\sqrt{x}-\sqrt{y}$$ must be positive. But this is not granted!!!
Why do you continue?

Try to apply your squaring approach to this:
For $$x$$ and $$y$$ positive, must $$x-y$$ be greater than $$\sqrt{x}+\sqrt{y}$$?
By the way, a math question (not on the GMAT), would never been formulated in this way - MUST BE...
It would simply read "Is $$x-y>\sqrt{x}+\sqrt{y}$$ for all positive $$x$$ and $$y$$?"
Would this wording of the question change your approach?
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Re: If x and y are positive, which of the following must be  [#permalink]

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26 Oct 2012, 20:15
Eva and Karishma,

Thanks for your detailed reply. I am sorry for late reply. You are correct in that one shouldn't conclude from A^2 > B^2 that A > B. That's not infer-able. I think that in this discussion, we are confusing "given" and "question" -- While solving or simplifying the inequality or any question , we have to assume that the equation/inequality holds good.

For instance, (given some condition); is it true that 4x + 4 > 3x ? The first step would be to simplify the equation to x>-4? Correct?

OR If the inequality requires cross multiplication, say, 4/x > 3/y; y>0; x!=0; Here, I would solve the equation with an assumption that x >0; I wouldn't assume that's THE ONLY possibility. I would also solve the equation with an assumption that x<0;

In the above question, I was trying to solve the equation and see under what conditions that equation holds good. It turns out, from our discussion, that the second inequality doesn't hold good. Let's assume that the inequality holds good i.e. xv>0. However, there is also an underlying assumption that x>y (we assumed that x-y> square of a number -- I think that this assumption is CRUCIAL which I missed initially. ); Hence, even though xy> 0 is true, that doesn't mean that xy>0 is a sufficient condition because the necessary condition x>y may not satisfied.

Here's an equation that's fun to solve.

2.5(x-2)> sqrt (x) [this is from GMATClub m05/q15] if-x-is-a-positive-integer-is-sqrt-x-2-5-x-5-1-x-91414.html

Method1: One of the methods is to plugin number.
MEthod2: Another would be to draw a line + parabolic curve. However, finding focus points etc would be difficult.
MEthod3:
Let's try by squaring. (Let's hold on to this)

Method4 :Let's first analyze the inequality to see what's going on:

Possible values of x : X<0 |X=0| 2>X>0 |2|X>2
X<0 => not possible
0<x<2 => (x is positive integer => x=1) => substitute in the equation => 1<2.5(1-2) => not possible.
X=2 => sqrt (2) ~1.4 <0 not possible
x>=3 => sqrt(3) < 2.5(3)-5 -> yes, true.

You see that by solving this inequality using analytical method, we can easily arrive at the answer even without looking at the two answer choices DS question! I didn't have to worry anything about primes etc.

Now, let's try my favorite square method (squaring an inequality is like an attempt to play blues chords on a major scale on your musical instrument--it's fun. Just as not all songs permit us to do that, we shouldn't do it. Now onward, I will refrain from squaring an inequality. I would probably try to do number plugging first. )
2.5(x-2) > sqrt(x)
6.25(x-2)^2 > x
Solving and re-arranging
x=2.65 or x=1.5; From our analysis, we know that 1.5 is not possible; hence, X> 2.65! This is also same as X>=3 (x = integer)

Thoughts?
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Re: If x and y are positive, which of the following must be  [#permalink]

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27 Oct 2012, 03:12
1
voodoochild wrote:
Eva and Karishma,

Thanks for your detailed reply. I am sorry for late reply. You are correct in that one shouldn't conclude from A^2 > B^2 that A > B. That's not infer-able. I think that in this discussion, we are confusing "given" and "question" -- While solving or simplifying the inequality or any question , we have to assume that the equation/inequality holds good.

For instance, (given some condition); is it true that 4x + 4 > 3x ? The first step would be to simplify the equation to x>-4? Correct?

OR If the inequality requires cross multiplication, say, 4/x > 3/y; y>0; x!=0; Here, I would solve the equation with an assumption that x >0; I wouldn't assume that's THE ONLY possibility. I would also solve the equation with an assumption that x<0;

In the above question, I was trying to solve the equation and see under what conditions that equation holds good. It turns out, from our discussion, that the second inequality doesn't hold good. Let's assume that the inequality holds good i.e. xv>0. However, there is also an underlying assumption that x>y (we assumed that x-y> square of a number -- I think that this assumption is CRUCIAL which I missed initially. ); Hence, even though xy> 0 is true, that doesn't mean that xy>0 is a sufficient condition because the necessary condition x>y may not satisfied.

Here's an equation that's fun to solve.

2.5(x-2)> sqrt (x) [this is from GMATClub m05/q15] if-x-is-a-positive-integer-is-sqrt-x-2-5-x-5-1-x-91414.html

Method1: One of the methods is to plugin number.
MEthod2: Another would be to draw a line + parabolic curve. However, finding focus points etc would be difficult.
MEthod3:
Let's try by squaring. (Let's hold on to this)

Method4 :Let's first analyze the inequality to see what's going on:

Possible values of x : X<0 |X=0| 2>X>0 |2|X>2
X<0 => not possible
0<x<2 => (x is positive integer => x=1) => substitute in the equation => 1<2.5(1-2) => not possible.
X=2 => sqrt (2) ~1.4 <0 not possible
x>=3 => sqrt(3) < 2.5(3)-5 -> yes, true.

You see that by solving this inequality using analytical method, we can easily arrive at the answer even without looking at the two answer choices DS question! I didn't have to worry anything about primes etc.

Now, let's try my favorite square method (squaring an inequality is like an attempt to play blues chords on a major scale on your musical instrument--it's fun. Just as not all songs permit us to do that, we shouldn't do it. Now onward, I will refrain from squaring an inequality. I would probably try to do number plugging first. )
2.5(x-2) > sqrt(x)
6.25(x-2)^2 > x
Solving and re-arranging
x=2.65 or x=1.5; From our analysis, we know that 1.5 is not possible; hence, X> 2.65! This is also same as X>=3 (x = integer)

Thoughts?

I think that in this discussion, we are confusing "given" and "question" -- While solving or simplifying the inequality or any question , we have to assume that the equation/inequality holds good.

I can speak just for myself, and I can assure you that I am not confusing between "given" and "question". It looks that you have major problems in distinguishing between solving an inequality and testing whether some inequalities hold for certain given values. With the cited question, you missed the point again.

I don't think there is anything left I can help you with.
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Re: If x and y are positive, which of the following must be  [#permalink]

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21 Jun 2014, 13:33
Bunuel, pls could u update OA and answer choices here? thanks!
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Re: If x and y are positive, which of the following must be  [#permalink]

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21 Jun 2014, 13:42
MensaNumber wrote:
Bunuel, pls could u update OA and answer choices here? thanks!

_____________
Done.
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Re: If x and y are positive, which of the following must be  [#permalink]

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29 Sep 2014, 02:47
noboru wrote:
If x and y are positive, which of the following must be greater than $$\frac{1}{\sqrt{x+y}}$$?

I. $$\frac{\sqrt{x+y}}{2}$$
II. $$\frac{\sqrt{x}+\sqrt{y}}{2}$$
III. $$\frac{\sqrt{x}-\sqrt{y}}{x+y}$$

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

III. We have
x+y+2√x√y>x+y ….. since x, y are positive, 2√x√y>0)
(√x+√y)^2>(√(x+y))^2
(√x+√y)>√(x+y)…taking square-root is valid since (√x+√y) and √(x+y) each is positive
(√x+√y)/(x+y)>√(x+y)/(x+y)..dividing by (x+y) is valid since (x+y) is positive
(√x+√y)/(x+y)>1/√(x+y)
Hence III must be greater than 1/√(x+y)
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Re: If x and y are positive, which of the following must be  [#permalink]

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29 Sep 2014, 02:50
Bunuel wrote:
prashantbacchewar wrote:
Problem looks weired. Please post the OA and official explaination.

Refer to the Walker's solution: it's fast (~30 sec) and elegant.

III. We have
x+y+2√x√y>x+y ….. since x, y are positive, 2√x√y>0)
(√x+√y)^2>(√(x+y))^2
(√x+√y)>√(x+y)…taking square-root is valid since (√x+√y) and √(x+y) each is positive
(√x+√y)/(x+y)>√(x+y)/(x+y)..dividing by (x+y) is valid since (x+y) is positive
(√x+√y)/(x+y)>1/√(x+y)
Hence III must be greater than 1/√(x+y)
Where am I wrong Bunuel?
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Re: If x and y are positive, which of the following must be  [#permalink]

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29 Sep 2014, 02:52
VeritasPrepKarishma wrote:
walker wrote:
Let's consider original statement: $$\frac{1}{\sqrt{x+y}}$$

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and $$\frac{1}{\sqrt{x+y}}$$ goes to infinity.

Nice approach... I think your first two lines given above are sufficient to answer the question. If the given expression $$\frac{1}{\sqrt{x+y}}$$ approaches infinity for some values of x and y, there is no way any other expression can be always greater than this expression. Some other expression can equal this expression by approaching infinity for the same values of x and y (e.g. the third expression), but cannot be greater than this.

So the moment one realizes that $$\frac{1}{\sqrt{x+y}}$$ tends to infinity, the answer is clear.

III. We have
x+y+2√x√y>x+y ….. since x, y are positive, 2√x√y>0)
(√x+√y)^2>(√(x+y))^2
(√x+√y)>√(x+y)…taking square-root is valid since (√x+√y) and √(x+y) each is positive
(√x+√y)/(x+y)>√(x+y)/(x+y)..dividing by (x+y) is valid since (x+y) is positive
(√x+√y)/(x+y)>1/√(x+y)
Hence III must be greater than 1/√(x+y)
Where am I wrong Karishma?
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Re: If x and y are positive, which of the following must be  [#permalink]

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04 Dec 2014, 09:22
walker wrote:
Let's consider original statement: $$\frac{1}{\sqrt{x+y}}$$

If I multiply the fraction by $${\sqrt{x+y}}$$ will I get $$\frac{ sqrt{x+y}}{\(x+y)}$$ ?
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Re: If x and y are positive, which of the following must be  [#permalink]

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04 Dec 2014, 10:18
1
joaogallegomoura wrote:
walker wrote:
Let's consider original statement: $$\frac{1}{\sqrt{x+y}}$$

If I multiply the fraction by $${\sqrt{x+y}}$$ will I get $$\frac{ sqrt{x+y}}{\(x+y)}$$ ?

If you multiply $$\frac{1}{\sqrt{x+y}}$$ by $${\sqrt{x+y}}$$ you get 1 ($$\frac{\sqrt{x+y}}{\sqrt{x+y}}=1$$).

You get $$\frac{ sqrt{x+y}}{\(x+y)}$$, if you multiply $$\frac{1}{\sqrt{x+y}}$$ by $$\frac{\sqrt{x+y}}{\sqrt{x+y}}$$:

$$\frac{1}{\sqrt{x+y}}*\frac{\sqrt{x+y}}{\sqrt{x+y}}=\frac{ sqrt{x+y}}{\(x+y)}$$.
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Re: If x and y are positive, which of the following must be  [#permalink]

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04 Dec 2014, 10:24
Bunuel wrote:
joaogallegomoura wrote:
walker wrote:
Let's consider original statement: $$\frac{1}{\sqrt{x+y}}$$

If I multiply the fraction by $${\sqrt{x+y}}$$ will I get $$\frac{ sqrt{x+y}}{\(x+y)}$$ ?

If you multiply $$\frac{1}{\sqrt{x+y}}$$ by $${\sqrt{x+y}}$$ you get 1 ($$\frac{\sqrt{x+y}}{\sqrt{x+y}}=1$$).

You get $$\frac{ sqrt{x+y}}{\(x+y)}$$, if you multiply $$\frac{1}{\sqrt{x+y}}$$ by $$\frac{\sqrt{x+y}}{\sqrt{x+y}}$$:

$$\frac{1}{\sqrt{x+y}}*\frac{\sqrt{x+y}}{\sqrt{x+y}}=\frac{ sqrt{x+y}}{\(x+y)}$$.

Great!

That was what I thought ... to multiply on both sides of the fraction

I think that by doing so it is easier to compare the fractions.

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Re: If x and y are positive, which of the following must be  [#permalink]

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17 Jan 2015, 08:49
Bunuel wrote:
mmcooley33 wrote:
walker wrote:
Let's consider original statement: $$\frac{1}{\sqrt{x+y}}$$

How can we approach the problem fast? Let's see when the original statement is very large : x,y ---> 0 and $$\frac{1}{\sqrt{x+y}}$$ goes to infinity.
Now, let's see what do our options at x,y ---> 0.

I) $$\frac{\sqrt{x+y}}{2}$$ goes to 0 at x,y ---> 0.

II) $$\frac{\sqrt{x}+\sqrt{y}}{2}$$ goes to 0 at x,y ---> 0.

III) $$\frac{\sqrt{x}-sqrt{y}}{x+y}$$ hm... it has x+y as a denominator. But what if x=y? At x=y it equals 0.

So, none of the options.

Can you explain how you used this in more detail, and perhaps give me some idea of how to apply this same principle going forward. Thanks a lot, i appreciate the help.

Just to add couple of words to Walker's great solution:

Note that we are asked "which of the following MUST be greater than $$\frac{1}{\sqrt{x+y}}$$?" not COULD be greater.

"MUST BE TRUE" questions:
These questions ask which of the following MUST be true (must be greater in our case), or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

As for "COULD BE TRUE" questions:
The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

So, if we find even one set of $$x$$ and $$y$$ for which $$\frac{1}{\sqrt{x+y}}$$ is greater than option I for example then it'll mean that option I is not ALWAYS greater then $$\frac{1}{\sqrt{x+y}}$$.

How can we increase the value of $$\frac{1}{\sqrt{x+y}}$$? Testing extreme examples: if $$x$$ and $$y$$ are very small (when their values approach zero) then denominator approaches zero and thus the value of the fraction goes to +infinity, becomes very large. In this case:

I. $$\frac{\sqrt{x+y}}{2}$$: nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming very small. So this option is not always more than given fraction;
II. $$\frac{\sqrt{x}+\sqrt{y}}{2}$$: the same here - nominator approaches zero, so the value of the whole fraction approaches zero as well, becoming very small. So this option is not always more than given fraction;

As for:
III. $$\frac{\sqrt{x}-\sqrt{y}}{x+y}$$, if $$x=y$$ then this fraction equals to zero and $$\frac{1}{\sqrt{x+y}}$$ has some value more than zero, so this option also is not always more than given fraction;

Answer: none of the options must be greater than the given fraction.

Hope it's clear.

Thanks Bunuel walker . Thanks for pointing out use of extreme cases for ruling out options.
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