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I Proof of an Inequality

  1. Dec 19, 2017 #1
    I'm trying to do some practice Putnam questions, and I'm stuck on the following:

    For ##a,b,c \geq 0##, prove that ##(a+b)(b+c)(c+a) \geq 8abc##


    I started off by expanding the brackets and doing some algebraic rearranging, but I don't think I've got anywhere.
    Should I be using induction? Or is this an algebraic problem?
    Thanks in advance
  2. jcsd
  3. Dec 19, 2017 #2


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    My hint is to first prove

    ##a + b \geq 2\sqrt {ab}##. You can start by using ##(a - b)^2 \geq 0##. If you prove it then see how you can leverage this for the whole expression you want to prove.

    EDIT: Your question is not about set theory, logic, probability or statistics so if it is not homework it should be posted in the "General Math" section.
    Last edited: Dec 19, 2017
  4. Dec 19, 2017 #3
    Apply AM-GM on ##2abc+a^2b+ac^2+b^2c+b^2a+bc^2+a^2c##.
  5. Dec 19, 2017 #4
    Thanks both.
  6. Dec 19, 2017 #5


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    Another common tactic is to start by dealing with the case where one or more variables is zero (i.e. lower bound of real non-negative expression is zero). After that you may assume all a,b, c are positive, then divide them out so all variables are on one side.

    In this case you'd take advantage of positivity and divide out ##abc## from each side. With a small bit of insight, you'll see that your problem 1 is just an n =3 case of problem 4.
  7. Dec 19, 2017 #6


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    I think ##3(a+b)(b+c)(c+a)=(a+b+c)^3-(a^3+b^3+c^3)## can be of use. At least one doesn't have to deal with so many terms.
  8. Dec 20, 2017 #7


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    Induction would work only if a,b,c were restricted to the domain of natural numbers and then again you would have to do triple nested induction. But here a,b,c are any positive real numbers.

    Just expand like @Buffu suggests and then use ##(x+y)\geq 2\sqrt{xy}## (already suggested by @QuantumQuest )for proper x and y for example ##x=a^2b,y=bc^2## e.t.c
    Last edited: Dec 20, 2017
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