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April 28th, 2015, 01:16 AM   #1
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a+b+c=3

If a+b+c=3
,(a,b&c>0)
then show that:
2(a²b+b²c+c²a)≥a²b²+b²c²+c²a²+3abc
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May 18th, 2015, 12:55 AM   #2
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At last I have found a counterexample: $a=0.89,\ b=2.10,\ c=0.01$.

LHS = 3.415198
RHS = 3.54975121
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May 18th, 2015, 03:27 AM   #3
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Quote:
Originally Posted by mared View Post
If a+b+c=3
,(a,b&c>0)
then show that:
2(a²b+b²c+c²a)≥a²b²+b²c²+c²a²+3abc
Command received, mared!
But not executed!
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Please show some kind of participation.
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May 18th, 2015, 04:01 AM   #4
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@prakhar and Olinguito

Your efforts are wasted. He never responds.

-Dan
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May 18th, 2015, 05:13 AM   #5
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He has replied with "send the solve" a few times
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