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Prove that (a+b)(b+c)(c+a) =/> 8abc

  1. Apr 24, 2010 #1
    prove that (a+b)(b+c)(c+a) =/> 8abc
    for all a,b,c =/> 0
    any1 pls.. thx.
     
  2. jcsd
  3. Apr 26, 2010 #2

    CompuChip

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    Re: inequalities

    Let's start by ordering them from largest to smallest: [itex]a \ge b \ge c[/itex].
    Then you can open the brackets and get 8 terms, two of which are precisely equal to abc. Picking one at random, say, b2c, can you show that this is larger than abc?
     
  4. Apr 26, 2010 #3

    mathman

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    Re: inequalities

    abc≥b2c, so the answer to your question is no.
     
  5. Apr 27, 2010 #4
    Re: inequalities

    You should use AM-GM inequality wich
    [tex]\sqrt{ab}[/tex] </= (a+b)/2
     
  6. Apr 27, 2010 #5
    Re: inequalities

    Given a,b,c =/> 0 implies
    a,b,c < 0 implies
    (a+b) < a
    (b+c) < b
    (c+a) < c
    .....
     
  7. Apr 27, 2010 #6

    Mentallic

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    Re: inequalities

    He was meant to say that =/> is read "equal or more".

    Compuchip's approach will be easiest.
     
  8. Apr 27, 2010 #7
    Re: inequalities

    When expanded you get

    [tex]a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^2 + abc + abc \geq 8abc[/tex]

    If you apply GM-AM inequality to the collection you get:

    [tex]a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^2 + abc + abc \geq 8\sqrt[8]{a^{8}b^{8}c^{8}}[/tex]
     
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