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 April 28th, 2015, 01:16 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 a+b+c=3 If a+b+c=3 ,(a,b&c>0) then show that: 2(a²b+b²c+c²a)≥a²b²+b²c²+c²a²+3abc
 May 18th, 2015, 12:55 AM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra At last I have found a counterexample: $a=0.89,\ b=2.10,\ c=0.01$. LHS = 3.415198 RHS = 3.54975121
May 18th, 2015, 03:27 AM   #3
Senior Member

Joined: Jul 2014
From: भारत

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Quote:
 Originally Posted by mared If a+b+c=3 ,(a,b&c>0) then show that: 2(a²b+b²c+c²a)≥a²b²+b²c²+c²a²+3abc
But not executed!
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Please show some kind of participation.

 May 18th, 2015, 04:01 AM #4 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,302 Thanks: 960 Math Focus: Wibbly wobbly timey-wimey stuff. @prakhar and Olinguito Your efforts are wasted. He never responds. -Dan
 May 18th, 2015, 05:13 AM #5 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 He has replied with "send the solve" a few times

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