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Math Help - Proving Inequalities

  1. #1
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    Proving Inequalities

    Prove that the inequality
    (a+7b)(4a+b) >= 3a^2 + 33ab + 2b^2
    holds for all real numbers a and b.
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  2. #2
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    OK, let's imagine that we're working on a piece of scrap paper, not writing out the finished solution. For inequalities you have to be careful when trying to reverse the steps.

    (a+7b)(4a+b) >= 3a^2 + 33ab + 2b^2
    <=>
    4a^2 + 29ab + 7b^2 >= 3a^2 + 33ab + 2b^2
    <=>
    a^2 - 4ab + 5b^2 >= 0
    <=>
    (a-2b)^2 + b^2 >= 0

    Now this last line is true, and each step in the chain is reversible. So we take our answer paper and write down
    (a-2b)^2 + b^2 >= 0
    expand
    hence a^2 - 4ab + 5b^2 >= 0
    add 3a^2 + 33ab + 2b^2 to both sides
    hence
    4a^2 + 29ab + 7b^2 >= 3a^2 + 33ab + 2b^2
    factorise LHS
    hence (a+7b)(4a+b) >= 3a^2 + 33ab + 2b^2

    Throw away the scrap paper and pretend that you thought of it all in that order!
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