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Problem with factorisation

  1. Mar 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Let a,b,c be positive real numbers. Prove that:

    a3+b3+c3≥a2b+b2c+c2a

    2. Relevant equations



    3. The attempt at a solution

    I assumed that a≥b≥c>0 following which I shifted the left side of this inequality to the right side giving

    a3+b3+c3-(a2b+b2c+c2a)≥0

    How do I do the required factorisation ... ???
     
  2. jcsd
  3. Mar 23, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    Are you sure that there is an easy way to factorize this?

    I would expect that this can be reduced to a problem similar to a/b + b/c + c/a > 3.
     
  4. Mar 23, 2013 #3
    I see a way to solve it, though it is a bit messy. First divide the entire equation by abc .
    A bit of algebra gets [itex] a^3+b^3+c^3 \geq ab + bc+ ac [/itex]. We can assume that a,b, and c are all greater than 1 since if (a,b,c) satisfies the inequality, then so does (ax,bx,cx) for any positive x. Using the fact that [itex] x^3 \geq x^2 if x>1 [/itex], it is therefore sufficient to show that
    [itex] a^2+b^2+c^2 \geq ab + bc +ac [/itex]

    Get everything over to the left hand side and multiply by 2. You will see that it factors nicely.

    Edit: Oops, disregard what I said, I made a mistake in my algebra.
     
    Last edited: Mar 23, 2013
  5. Mar 23, 2013 #4
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