Suppose

then

Please tell me how did you get that?

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

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

]]>then

]]>

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

]]>Please solve the problem for me

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

]]>Where a, b and c are distinct real numbers.]]>