Prove a/sqrt(b+c) + c/sqrt(b+a) + c/sqrt(a+b) > 0 for positive a,b,c
Not sure if this is even correct, but i saw this somewhere long ago.
Edit: I'm thinking CauchySchwarz right now.
Almost Nesbitt's inequality
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Almost Nesbitt's inequality
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Re: Almost Nesbitt's inequality
Um, well, if you copied it correctly, then this is almost trivially true. The result of the square root function is always positive (with positive arguments), and since a, b and c are given as positive, then you are saying that the sum of three positive fractions is greater than zero...surprise surprise.
Or was there a geometric interpretation or something that I'm missing?
Answer: There isn't. But the question may have been miscopied.</bluntness>
Or was there a geometric interpretation or something that I'm missing?
Answer: There isn't. But the question may have been miscopied.</bluntness>
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Re: Almost Nesbitt's inequality
No, looks like i just have it incorrect... I do remember some kind of twist on nesbitt's though, involving radicals... but that means this is not it.
GENERATION 705  992 i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.

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Re: Almost Nesbitt's inequality
Is there still an inequality if we remove the bounds for a,b,c ?
GENERATION 705  992 i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.
Re: Almost Nesbitt's inequality
Okay, fixed that (and the other odd bit which I assume was a typo.)Paradoxica wrote:Prove a/sqrt(b+c) + c/sqrt(b+a) + b/sqrt(a+c) > 0 for positive or negative a,b,c
Is this inequality true now? Much more interesting question. Let's see....
...Seems like we're going to have some complex numbers on our hands. The inequality relation isn't defined for complex numbers, is it?
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Re: Almost Nesbitt's inequality
no. what if you replace it with the euclidean distance formula? i suppose that'll still be positive.
this isn't the inequality i'm after. the one i remember had a nonzero constant on the other side...
this isn't the inequality i'm after. the one i remember had a nonzero constant on the other side...
GENERATION 705  992 i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.
Re: Almost Nesbitt's inequality
Wildcard wrote:Okay, fixed that (and the other odd bit which I assume was a typo.)Paradoxica wrote:Prove a/sqrt(b+c) + c/sqrt(b+a) + b/sqrt(a+c) > 0 for positive or negative a,b,c
Is this inequality true now? Much more interesting question. Let's see....
...Seems like we're going to have some complex numbers on our hands. The inequality relation isn't defined for complex numbers, is it?
If you add the restriction that a + b > 0, b + c > 0, and a + c > 0, so that the squareroots are never complex, then I believe the statement is true.
If all a, b, and c are positive, then the statement is trivially true. If two or more of a, b, and c are negative, then one of the conditions is violated. So only one can be negative, WOLG I will call it a. Then b > a and c > a, and b + c > a + b and b + c > a + c. Then c/sqrt(b + a) > a/sqrt(b + c) and b/sqrt(a + c) > a/sqrt(b +c). Therefore a/sqrt(b+c) + c/sqrt(b+a) + b/sqrt(a+c) > 0.

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Re: Almost Nesbitt's inequality
how about a/sqrt(b^2 + c^2) and it's permutations
also, the same as above, but without the radicals
what will be on the other side of the inequalities if we allow all real values of a,b and c?
also, the same as above, but without the radicals
what will be on the other side of the inequalities if we allow all real values of a,b and c?
GENERATION 705  992 i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.

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Re: Almost Nesbitt's inequality
I think for my first post, the original question was to find the lower bound of the expression, assuming a, b, c are all strictly positive.
GENERATION 705  992 i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.
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