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PS Fraction

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Senior Manager
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PS Fraction [#permalink]

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New post 24 Aug 2008, 06:34
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Re: PS Fraction [#permalink]

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New post 24 Aug 2008, 06:42
I posted this solution to beatthegmat, so I'll just copy-paste here:

Fractions become easier to compare when we have the same numerator or the same denominator. Notice that 1/root(x+y) = root(x+y)/(x+y).

We want to know if any of I, II or III must be larger than root(x+y)/(x+y). Note that this must be positive, while III can be negative, so III does not need to be larger.

I has the same numerator, and it would be larger if its denominator, 2x, was certain to be smaller than x+y, i.e. if 2x < x+y, or x < y. We don't know whether x is smaller than y, so I does not need to be larger than root(x+y)/(x+y).

II has the same denominator as root(x+y)/(x+y). We thus want to know which has the larger numerator. That is, if we knew that

root(x+y) < root(x) + root(y)

was always true for positive x and y, we'd know that II must be larger than the quantity in the question. Well,

[root(x+y)]^2 = x+y < x + 2root(xy) + y = [root(x) + root(y)]^2

so II must be larger. Or you can look at it this way:

Of course we can make a right-triangle with legs of length root(x) and root(y). By Pythagoras, the hypotenuse will be root(x+y). But the sum of two sides of a triangle is always greater than the third side:

root(x) + root(y) > root(x+y)

which is what we were trying to prove.
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Kudos [?]: 2004 [0], given: 6

Senior Manager
Senior Manager
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Joined: 19 Mar 2008
Posts: 351

Kudos [?]: 68 [0], given: 0

Re: PS Fraction [#permalink]

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New post 24 Aug 2008, 07:24
Thanks, Ian

New rule I learnt:
root(x+y) < root(x) + root(y)

was always true for positive x and y,

+ 1 for you

Kudos [?]: 68 [0], given: 0

Re: PS Fraction   [#permalink] 24 Aug 2008, 07:24
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