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Agnishom
2013-09-23 13:01:37

#### gAr wrote:

Suppose

then

Please tell me how did you get that?

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

anonimnystefy
2013-09-20 05:47:37

Hi gAr

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

gAr
2013-09-20 02:52:56

Suppose

then

anonimnystefy
2013-09-19 02:56:57

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.

bobbym
2013-09-19 02:00:19

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.

Agnishom
2013-09-19 01:46:54

They are distinct.

Please solve the problem for me

bobbym
2013-09-19 01:36:56

You can find solutions if a = b.

Agnishom
2013-09-19 01:31:53

then??

bobbym
2013-09-19 01:12:13

I do not think so.

Agnishom
2013-09-19 00:37:59

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

bobbym
2013-09-19 00:13:01

Agnishom
2013-09-19 00:04:25

Is it possible that

Where a, b and c are distinct real numbers.