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24 Aug 2008, 06:34
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24 Aug 2008, 06:42
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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.
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.
24 Aug 2008, 07:24
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Thanks, Ian
New rule I learnt:
root(x+y) < root(x) + root(y)
was always true for positive x and y,
+ 1 for you
New rule I learnt:
root(x+y) < root(x) + root(y)
was always true for positive x and y,
+ 1 for you
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
