Arithmetic

Surely the answer cannot include answer (3)?
Because, if x = y = 1
then answer (3) could give

$$\sqrt{1}-\sqrt{1}\ =\ 0\ <\ \frac{1}{\sqrt{1\ +\ 1}}\ for\ positive\ values\ of\ \ roots$$

The answer must be NONE [A], because all answers include square roots which could generate a negative answer.
As $$\sqrt{x}\ =\ +\ or\ -\ \sqrt{x}$$
Therefore the "MUST" statement is contravened.

Roland2rule wrote: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

Let x=1 and y=1.
1/√(x+y) = 1/√(1+1) = 1/√2.
Eliminate any expression not greater than 1/√2.

Scanning options I, II and III, we can quickly see that the numerator of III is equal to 0:
(√x-√y)/(x+y) = (√1-√1)/(1+1) = 0.
Eliminate any answer choice that includes III (D and E).

I: √(x+y)/2x
√(1+1)/(2*1) > 1/√2
√2/2 > 1/√2.
Cross-multiplying, we get:
√2 *√2 > 2*1
2>2.
Doesn't work.
Eliminate any remaining answer choice that includes I (B).

To compare fractions:
Multiply the denominator in each fraction by the NUMERATOR in the OTHER fraction.
The numerator that yields the greater product belongs to the greater fraction.

Multiplying the denominator in II by the numerator in the prompt, we get:
(x+y)(1) = x+y.
Multiplying the denominator in the prompt by the numerator in II, we get:
√(x+y) * (√x + √y)

= √(x² + xy) + √(xy + y²)

= √(more than x²) + √(more than y²)

= (more than x) + (more than y).

Since the product in blue is greater than the product in red, the numerator in II yields the greater product.
Thus, the fraction in II must be greater than the fraction in the prompt.
Eliminate any remaining answer choice that does not include II (A).

The correct answer is C.