inequality proof
- Thread starter VelvetSky
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a,b,c are reals,positive reals . a,b,c are in the interval (0 , +infinity), R*+
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- Apr 12, 2005
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a = b = c = w is easily demonstrated.
Can you demonstrate that this is the minimum value attained by the left-hand expression, when all three are equal?
How about ordering them, in a general way.
1) Pick a, b, and c
2) Sort them so that a < b = a + n < c = a + m = b + q, where n, m, and q >= 0
Taking just a and b, we have:
\(\displaystyle \frac{a^{2}+b^{2}}{a+b} = \frac{a^{2}+(a+n)^{2}}{a+a+n} = \frac{2a^{2}+2an+n^2}{2a+n} = a + \frac{n}{2} + \frac{n^{2}}{2\cdot(2a+n)}> a\) unless n = 0
I think we're done.
There's probably a beautiful way to do it. This wasn't it.
Can you demonstrate that this is the minimum value attained by the left-hand expression, when all three are equal?
How about ordering them, in a general way.
1) Pick a, b, and c
2) Sort them so that a < b = a + n < c = a + m = b + q, where n, m, and q >= 0
Taking just a and b, we have:
\(\displaystyle \frac{a^{2}+b^{2}}{a+b} = \frac{a^{2}+(a+n)^{2}}{a+a+n} = \frac{2a^{2}+2an+n^2}{2a+n} = a + \frac{n}{2} + \frac{n^{2}}{2\cdot(2a+n)}> a\) unless n = 0
I think we're done.
There's probably a beautiful way to do it. This wasn't it.