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

If x and y are positive, which of the following must be greater than \(\frac{1}{\sqrt{x+y}}\)?

1. \(\frac{\sqrt{x+y}}{2x}\)

2. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\)

3. \(\frac{\sqrt{x}-\sqrt{y}}{x+y}\)

(A) None
(B) 1 only
(C) 2 only
(D) 1 and 3 only
(E) 2 and 3 only

First of all \(\frac{1}{\sqrt{x+y}}\) is always positive. This by the way eliminates option III right away as \(\sqrt{x}-\sqrt{y}\) (numereator) may or may not be positive, so we should concentrate on I and II

Next:

\(\sqrt{x}+\sqrt{y}\) is always greater than \(\sqrt{x+y}\) (well in fact if both \(x\) and \(y\) are 0, they are equal but it's not the case as given that \(x\) and \(y\) are positive). To check this: square them \((\sqrt{x}+\sqrt{y})^2=x+2\sqrt{xy}+y>x+y=\sqrt{x+y}^2\)

Let's proceed:

SOLUTION #1
\(\frac{1}{\sqrt{x+y}}=\frac{\sqrt{x+y}}{x+y}\)

I. \(\frac{\sqrt{x+y}}{2x}\) --> nominators are the same, obviously denominator \(2x\) may or may not be greater than \(x+y\). OUT.

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) --> denominators are the same and nominator \(\sqrt{x}+\sqrt{y}\) (as we've already discussed above) is always greater than \(\sqrt{x+y}\). OK

III. Well, we could even not consider this one as our expression \(\frac{1}{\sqrt{x+y}}\) is always positive and the \(\sqrt{x}-\sqrt{y}\) (numerator) can be negative. OUT

Answer C.


SOLUTION #2
The method called cross multiplication:
Suppose we want to know which positive fraction is greater \(\frac{9}{11}\) or \(\frac{13}{15}\): crossmultiply \(9*15=135\) and \(11*13=143\) --> \(135<143\) which fraction gave us numerator for bigger value 143? \(\frac{13}{15}\)! Thus \(\frac{13}{15}>\frac{9}{11}\).

Lets do the same with our problem:
I. \(\frac{\sqrt{x+y}}{2x}\) and \(\frac{1}{\sqrt{x+y}}\) --> \(\sqrt{x+y}*\sqrt{x+y}=x+y\) and \(2x*1=2x\). \(x+y\) may or may not be greater than \(2x\). OUT

II. \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) and \(\frac{1}{\sqrt{x+y}}\) --> \((\sqrt{x}+\sqrt{y})(\sqrt{x+y})\) and \(x+y\). Divide both sides by \(\sqrt{x+y}\) --> \(\sqrt{x}+\sqrt{y}\) and \(\sqrt{x+y}\). We know that \(\sqrt{x}+\sqrt{y}\) is always greater, which one gave the numerator for it: \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\), so \(\frac{\sqrt{x}+\sqrt{y}}{x+y}\) is always greater than \(\frac{1}{\sqrt{x+y}}\). OK

III. Well, we could even not consider this one as our expression \(\frac{1}{\sqrt{x+y}}\) is always positive and the \(\sqrt{x}-\sqrt{y}\) (numerator) can be negative. OUT

Answer C.

Hope it's clear.