Advanced Algebra Prob: If x, y > 0, which must be greater

[artshep]

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

a. None
b. 1 only
c. 2 only
d. 1 and 3
e. 2 and 3

The answer to the question is (c); however, I am unable to see how this is the correct solution.

If I take the term in (2), which is the term referred to in answer choice (c) and take the SQRT of both the denominator and the numerator, I get:

. . .SQRT (SQRT(x) + SQRT(y)) / (SQRT(x+y))

This now has the same denominator as the subject term. However, for this to be always greater than the subject term, the numerator has to always be greater than 1. This is not necessarily the case. For example, if x and y are both less than 1...say 1/2 for example...then this term will not be greater than the subject term.

Can anyone find where I am making a mistake in my approach and if so suggest how to correct my approach?

Thanks,
Artis

 

[Denis]

"If x and y are positive, which of the following must be greater than 1/(SQRT(x+y)) ?
(2) (SQRT(x) + SQRT(y)) / (x + y) "

Assume they're equal:
(SQRT(x) + SQRT(y)) / (x + y) = 1 / SQRT(x + y) ; square both sides:

[x + 2SQRT(xy) + y] / (x + y)^2 = 1 / (x + y)

[x + 2SQRT(xy) + y] / (x + y) = 1

Play with that....

 

[stapel]

You have:

. . . . .\(\displaystyle \L \frac{1}{\sqrt{x\, +\, y}\)

...with x, y > 0. Considering the three given expressions, and forming differences, we get:

. . . . .\(\displaystyle \L \frac{\sqrt{x\, +\, y}}{2x}\, -\, \frac{1}{\sqrt{x\, +\, y}}\, =\, \frac{(x\, +\, y)\, -\, 2x}{\sqrt{2x}\, \sqrt{x\, +\, y}}\, =\, \frac{y\, -\, x}{\sqrt{2x}\, \sqrt{x\, +\, y}}\)

By definition, the square roots are positive. But we cannot assume that y is greater than x, so we cannot conclude that y - x, and thus the entire fraction, is positive. Since this difference might be negative, then the subtracted term might be larger than (1). So we cannot conclude that (1) will always be greater.

Turning to (3), we get:

. . . . .\(\displaystyle \L \frac{\sqrt{x}\, -\, \sqrt{y}}{x\, +\, y}\, -\, \frac{1}{\sqrt{x\, +\, y}}\, =\, \frac{\sqrt{x}\, -\, \sqrt{y}\, -\, \sqrt{x\, +\, y}}{x\, +\, y}\)

If x < 1 < y, then the numerator will be negative, showing an example of where (3) could be smaller than the subtracted term.

Turning at last to (2), we get:

. . . . .\(\displaystyle \L \frac{\sqrt{x}\, +\, \sqrt{y}}{x\, +\, y}\, -\, \frac{1}{\sqrt{x\, +\, y}}\, =\, \frac{\sqrt{x}\, +\sqrt{y}\, -\, \sqrt{x\, +\, y}}{x\, +\, y}\)

For this to be non-positive, you would have to have the numerator zero or negative. What do you get if you set the numerator equal to zero?

Eliz.

 

[artshep]

Thanks...

Stapel - thanks for the explanations. That is a great way to go about solving the problem in total. Answering your question, for the numerator to be equal or less than zero, x or y would have to equal zero, which is impossible according to the given information.

Denis - thanks for the tip on setting that up.