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

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.

Karishma,
Thanks for your reply. Your replies are always helpful. However, I have a different opinion for two reasons:

#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.

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

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)

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?

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.

Bunuel, pls could u update OA and answer choices here? thanks!

_____________
Done.

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)
Correct answer C.

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)
Correct answer C.
Where am I wrong Bunuel?

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)
Correct answer C.
Where am I wrong Karishma?

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.

Many thanks for your clarification.

In "MUST BE TRUE" questions, you need to use POE and be ready to prove that none of the options/statements are possible. Even 1 not possible scenario will make that option not allowed.

Had this question been "could be true" instead, II only would have been correct with x=y=1. But as this is a MUST BE TRUE question, you need to make sure to find 1 set of (x,y) that will negate the given conditions.

i) and iii) can be easily POE-d by assuming x=y=1. In both these cases the resulting values will NOT BE GREATER than \(1/(x+y)^{0.5}\).

For ii), you can clearly see x=y=1 gives you a value greater than \(1/(x+y)^{0.5}\) but what about x=y=4 again you get a value greater. So lets take x=y=0.25. In this case you will end up getting ii) < \(1/(x+y)^{0.5}\). Hence this expression as well is NOT always true and is hence eliminated.

As you eliminated all the 3 possible options, OA must be E (none).

Hope this helps.