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Is it possible that

Where a, b and c are distinct real numbers.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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I would also say no, because if it was yes Suman Sir would not have asked it.

I guess we will get to see some thing like sin θ > 2 as a result which is a contradiction

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bobbym****Administrator**- From: Bumpkinland
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I do not think so.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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then??

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**bobbym****Administrator**- From: Bumpkinland
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You can find solutions if a = b.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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They are distinct.

Please solve the problem for me

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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That is what I mean, the numerator is only equal to the denominator when a=b=c.

When a,b,c>0 it is easy to prove that fraction is greater than 1 therefore there can be no sin(θ).

This is equal to,

There is only equality when a=b=c which violates the given conditions. That completes the proof for a,b,c>=0

Or use the AMGM.

Now add up the 3 inequalities and divide by 2 and the result follows. There is only equality when a=b=c.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,518

Hi Agnishom

Start with the known inequality (a-b)^2+(b-c)^2+(c-a)^2>0 for distinct real numbers a, b and c.

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**gAr****Member**- Registered: 2011-01-09
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Suppose

then

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Hi gAr

That doesn't work for ab+bc+ac<0, but that case is done similarly.

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gAr wrote:

Suppose

then

Please tell me how did you get that?

Later edit: Sorry, got it from bobbym's last post

*Last edited by Agnishom (2013-09-22 15:41:33)*

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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