yvonne12 wrote:If Xand Y are positive, which of the following would be greater than 1/ sqrt x+y?

1. Sqrt x+y/2x

2. sqrt x + sqrt y/ x+y

3. Sqrt X - sqrt y /x+y

I think this is best solved by subst values for x and y.

Take x = 1, y = 8

1/sqrt x+y = 1/3 which is < 1

1 - sqrt (9/2) > 1

2 - sqrt (1) + sqrt (8/9) > 1

3 - sqrt (1) - sqrt (8/9) = 1 - 2*1.4/3 = .2/3
This is less than 1/3


I think 1 and 2 alone satisfy the conditions.

Jay

What is the difference between

"Which of them would be greater" and "Which of them must be greater"

f2001290 wrote:Jay

What is the difference between

"Which of them would be greater" and "Which of them must be greater"

In the context of the question, I guess "would be" and "must be"
have the same meaning.

In general "must be" is a condition that should ALWAYS be true.

Jay

OA for this question is 2. Refer the attachment. I got it from some other forum.

f2001290 wrote:Jay

OA for this question is 2. Refer the attachment. I got it from some other forum.

Since we are asked prove "must", we have to ensure that the condition
holds true for all values.

Looking at I, II and III, you can infer that calculations will involve a lot
of root.

To simplify, Take a pythogorean triplet, x=64, y=36, so that x+y = 100

I and III immediately fall apart. You can eliminate B, D and E.

To evaluate II,

sqrt-x + sqrt-y / (x + y)
= (sqrt-x + sqrt-y) / sqrt (x+y)*sqrt(x+y) ..... eqn-a

To see if II is greater than 1/sqrt(x+y), we have to prove

sqrt-x + sqrt-y / sqrt(x+y) > 1 i.e. numerator of eqn-a should be > 1

i.e. prove sqrt-x + sqrt-y > sqrt(x+y)

Since we are dealing with positive values for x and y, we know
the above will hold true. Hence sufficient.

So, II it is.

Is there any methodical approach to this problem.

At times, I feel that substituting values is a tedious job and there is ample scope for getting such inequality problems wrong.