Those Radicals......

GHong14 wrote:which of the following must be greater than 1/ sq(x+Y)

I. sq(x+y)/2x
II. sq(x)+sq(y)/(x+y)
III. sq(x)-sq(y)/(x+y)

In addition would using difference of squares in this problem help solve it?

the difference doesn't help me too much here.
one way I solve this problem is to select smart numbers for x and y so that we can return sqr.roots,
say x=16 and y=9 or the other way around (doesn't really matter, can be anything)

another approach is to switch the fractions (here we may need to plug in 9 and 16 still)

1/ sq(x+Y) and sq(x+y)/2x compare with 1*2x and sq(x+Y)*sq(x+y)

and so on ...

by plug in, I have only the second [ II. sq(x)+sq(y)/(x+y) ] satisfies.

Night reader wrote:
the difference doesn't help me too much here.
one way I solve this problem is to select smart numbers for x and y so that we can return sqr.roots,
say x=16 and y=9 or the other way around (doesn't really matter, can be anything)

another approach is to switch the fractions (here we may need to plug in 9 and 16 still)

1/ sq(x+Y) and sq(x+y)/2x compare with 1*2x and sq(x+Y)*sq(x+y)

and so on ...

by plug in, I have only the second [ II. sq(x)+sq(y)/(x+y) ] satisfies.

X and Y can be in the form of 1/2 or 1/3 In that case picking numbers will be very tedious.

I think there must be a particular approach to solve this one, ??????????????

goyalsau wrote:

Night reader wrote:
the difference doesn't help me too much here.
one way I solve this problem is to select smart numbers for x and y so that we can return sqr.roots,
say x=16 and y=9 or the other way around (doesn't really matter, can be anything)

another approach is to switch the fractions (here we may need to plug in 9 and 16 still)

1/ sq(x+Y) and sq(x+y)/2x compare with 1*2x and sq(x+Y)*sq(x+y)

and so on ...

by plug in, I have only the second [ II. sq(x)+sq(y)/(x+y) ] satisfies.

X and Y can be in the form of 1/2 or 1/3 In that case picking numbers will be very tedious.

I think there must be a particular approach to solve this one, ??????????????

ask experts? it's beyond my current math preparation ...
I pick numbers and simplify further if fractions are involved

GHong14 wrote:which of the following must be greater than 1/ sq(x+Y)

I. sq(x+y)/2x
II. sq(x)+sq(y)/(x+y)
III. sq(x)-sq(y)/(x+y)

Hi there!

I. Please note that (*) sq(x+y)/2x > 1/sq(x+y) is equivalent to x+y > 2x whenever x is greater than zero, because if x>0, from the fact that we MAY (and should) consider sq(x+y) positive (it is always non-negative and it is a denominator in the question stem), multiplying both sides of the inequality (*) by the POSITIVE factor 2*x*sq(x+y) you get what I said.

Therefore from the fact that x+y > 2x is equivalent to y > x, we understand that I. would be "yes" if we could guarantee that y > x > 0. This is not guaranteed, therefore it is easy to find a counter-example:

Take x = 2 and y = 1, then sq(3)/4 is not greater than 1/sq(3) for sure! (No need for calculations if you understand what I did above.)

(Of course we need NOT consider x < 0, because in the case x > 0 we already found a counter-example.)

> IMPORTANT: do you see how "math" gives you the path to look for counter-examples? That´s the key of problems like this one!!

Before I solve the other items, please check whether II. is as mentioned or if it should be (sq(x)+sq(y))/(x+y) , please. The same with III, please.

Regards,
Fabio.

P.S.: I will take approx. 4 hours to come back here, because I have to teach some students "in a row", now!

fskilnik wrote: Before I solve the other items, please check whether II. is as mentioned or if it should be (sq(x)+sq(y))/(x+y) , please. The same with III, please.

Take x = 0 and y = 1 as a counter-example to ANY of the two "II. versions" and ANY of the two "III. versions"!!

> Important: when "no math" seems to simplify the question "involved", try TRIVIAL (viable) examples to look for "potential" counter-examples...

Regards,
Fabio.

Rahul@gurome wrote:But we can assume that x and y are positive as the denominator of some terms are x or (x + y) and some term in the expressions contains √x and √y.

Nope. We can assume that sq(*) has * as positive or null, just that, for every expression given in the question. When the expression is in the denominator, we can also assume that * is positive, for sure.

Rahul@gurome wrote:√(xy) > 0 is always true as x and y are not zero. Thus (II) is always greater than 1/√(x + y).

The conclusion is false, because sq(xy) may be zero if y is zero (as I took for counter-example). Conditions given in the question stem allows us to guarantee that BOTH x and y cannot be SIMULTANEOUSLY zero but please note that y = 0 and x positive well-defines the expression given at the question itself AND all expressions given in I, II and III.

Therefore the answer is NONE of the statements I, II, III is necessarily greater than the expression given.

Regards,
Fabio.

P.S.: if this question was taken from some source where the "OA" is different from what I said, please show me where my reasoning is flawed or accept the fact that "OA" are created by human beings, that means nobody is perfect...

Hi Rahul!
Just technology could make two GMAT teachers from India and Brazil to talk as we were in the same place, isn´t it ?

Rahul@gurome wrote:But I assumed (and declared initially) x and y both as positive and approached accordingly.

Yes... IF your assumption could be inferred by the question itself, implicitly or explicitly, you would be right.

Rahul@gurome wrote:The question should contain some declaration about the properties of the variables
because assuming number properties of the variables from the options given is not a legitimate way of solving a problem.

When you are asked "1/x > 0 ?" you may assume that 1/x is defined, that is, x is not zero. Here is exactly the same thing, but we have plenty of "implicit conditions" because we have plenty expressions in the question stem (what includes statements I, II and III). In other words, I guess this problem has nothing illegitimate (neither in the question itself, nor in my solution).

Regards,
Fabio.