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Almost Nesbitt's inequality

Posted: Wed May 20, 2015 7:37 am UTC
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 Cauchy-Schwarz right now.

Re: Almost Nesbitt's inequality

Posted: Wed May 20, 2015 7:41 am UTC
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>

Re: Almost Nesbitt's inequality

Posted: Wed May 20, 2015 7:47 am UTC
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.

Re: Almost Nesbitt's inequality

Posted: Wed May 20, 2015 7:49 am UTC
Is there still an inequality if we remove the bounds for a,b,c ?

Re: Almost Nesbitt's inequality

Posted: Wed May 20, 2015 8:07 am UTC
Paradoxica wrote:Prove a/sqrt(b+c) + c/sqrt(b+a) + b/sqrt(a+c) > 0 for positive or negative a,b,c
Okay, fixed that (and the other odd bit which I assume was a typo.)

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?

Re: Almost Nesbitt's inequality

Posted: Wed May 20, 2015 8:52 am UTC
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 non-zero constant on the other side...

Re: Almost Nesbitt's inequality

Posted: Wed May 20, 2015 10:24 pm UTC
Wildcard wrote:
Paradoxica wrote:Prove a/sqrt(b+c) + c/sqrt(b+a) + b/sqrt(a+c) > 0 for positive or negative a,b,c
Okay, fixed that (and the other odd bit which I assume was a typo.)

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.

Re: Almost Nesbitt's inequality

Posted: Wed May 20, 2015 11:52 pm UTC
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?

Re: Almost Nesbitt's inequality

Posted: Wed Jun 03, 2015 11:00 pm UTC
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.