If x and y are positive, which of the following must be greater than 1/√(x+y)?
I. √(x+y) / 2x
II. (√x + √y) / (x+y)
III. (√x - √y) / (x+y)
A. None
B. I only
C. II only.
D. I and III
E. II and III
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 (C and E).
I: √(x+y)/2x > 1/√2
√(1+1)/(2*1) > 1/√2
√2/2 > 1/√2.
Cross-muliplying, we get:
√2 *√2 > 2*1
2>2.
Doesn't work.
Eliminate any remaining answer choice that includes (B).
II: (√1 + √1) / (1+1) > 1/√2
(√1 + √1) / (1+1) > 1/√2
2/2 > 1/√2
1 > 1/√2.
This works.
Test whether (√x+ √y) / (x+y) > 1/√(x+y) if x and y are far apart.
Let x=1 and y=100.
II: (√1 + √100) / (1+100) > 1/√(1+100)
(1+10)/101 > 1/√101.
Since √101 ≈ 10, we get:
11/101 > 1/10.
Cross-multiplying, we get:
10*11 > 1*101
110 > 101.
Still works.
The correct answer is
B.
Proof for statement II:
Rephrase 1/√(x+y) to have the same denominator as statement II:
1/√(x+y) = [1 * √(x+y)] / [√(x+y) * √(x+y)] = √(x+y) / (x+y).
Comparing statement II with the resulting expression above, we get:
(√x + √y) / (x+y) > √(x+y) / (x+y)
(√x + √y) > √(x+y)
(√x + √y)² > √(x+y)²
x + 2√x√y + y > x+y
2√x√y > 0.
Since x and y are positive, the resulting inequality MUST BE TRUE.
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