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.
]]>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
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