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?

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.



Here your question is:
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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#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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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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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?