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Is it possible that

Where a, b and c are distinct real numbers.

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

I'm not crazy, my mother had me tested.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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

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

I'm not crazy, my mother had me tested.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I do not think so.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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then??

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

I'm not crazy, my mother had me tested.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

You can find solutions if a = b.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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'

I'm not crazy, my mother had me tested.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Suppose

then

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

Hi gAr

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

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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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'

I'm not crazy, my mother had me tested.

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