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 Posts: 1,178 Thanks: 230 | |

May 18th, 2015, 04:01 AM |
#4 |

Math Team Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 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 |