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Math Help - Inequality involving absolute value

  1. #1
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    Inequality involving absolute value

    If a^2+b^2=2abc, prove that |c|\geq a and |c|\geq b

    I run out of ideas.

    I tried a^2+b^2+2ab=2ab+2abc
    I tried a^2=2abc-b^2

    Does anyone know how to prove this?
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  2. #2
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    Quote Originally Posted by novice View Post
    If a^2+b^2=2abc, prove that |c|\geq a and |c|\geq b

    I run out of ideas.

    I tried a^2+b^2+2ab=2ab+2abc
    I tried a^2=2abc-b^2

    Does anyone know how to prove this?
    It's good that you can't prove this because it is not true! Take a= b= 3 and c= 1. Then a^2+ b^2= 9+9 18= 2(3)(3)(1)= 2abc but |c|= 1< a= 3 and |c|= 1< b= 3.
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  3. #3
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    Quote Originally Posted by novice View Post
    If a^2+b^2=2abc, prove that |c|\geq a and |c|\geq b

    I run out of ideas.

    I tried a^2+b^2+2ab=2ab+2abc
    I tried a^2=2abc-b^2

    Does anyone know how to prove this?
    That result is not true. The result should be that |c|\geqslant1. For example, if a=3 and b=4 then c = 25/24, which is greater than 1, but less than a or b. Edit: HallsofIvy beat me to it!

    To see why |c|\geqslant1, divide the equation by b^2 (assuming that b\ne0). It becomes \bigl(\tfrac ab\bigr)^2 + 1 = 2\bigl(\tfrac ab\bigr)c. Now put x = \tfrac ab, so that x^2-2cx+1=0. The condition for that quadratic to have real roots is |c|\geqslant1.
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  4. #4
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    Thank you both, HallsofIvy and Opalg. It's very nice to have different ways of seeing things. I sure have a lot to learn still.
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  5. #5
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    If a^2 + b^2 = R^2,
    2abc = R^2

    then your inequality holds.
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