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Math Help - Inequality proof

  1. #1
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    Inequality proof

    show that

    ab+ac+bc=<a^2+b^2+c^2

    i've done a similar example and trying to do what i did there with this one but just not working out. below is the one i've already done

    2ab=<a^2+b^2
    i did like this

    since

    (a-b)^2=a^-2ab+b^2
    thus
    a^2+b^2=(a-b)^2+2ab
    but
    (a-b)^2>=0
    then
    a^2+b^2>=2ab
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  2. #2
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    Quote Originally Posted by action259 View Post
    show that

    ab+ac+bc=<a^2+b^2+c^2

    i've done a similar example and trying to do what i did there with this one but just not working out. below is the one i've already done

    2ab=<a^2+b^2
    i did like this

    since

    (a-b)^2=a^-2ab+b^2
    thus
    a^2+b^2=(a-b)^2+2ab
    but
    (a-b)^2>=0
    then
    a^2+b^2>=2ab
    Sounds like a good start. Try this:
    (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc

    If you haven't seen this before, I recommend you multiply out (a + b + c)(a + b + c) for yourself.

    -Dan
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  3. #3
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    ok so i get this
    (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc
    then
    (a + b + c)^2 - 2ab - 2ac - 2bc = a^2 + b^2 + c^2
    and
    (a + b + c)^2>=0
    so
    - 2ab - 2ac - 2bc =< a^2 + b^2 + c^2
    which is close to what i'm trying to show just have the multiple of 2 and negative signs so not sure this will work
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  4. #4
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    Quote Originally Posted by action259 View Post
    show that

    ab+ac+bc=<a^2+b^2+c^2
    You can use the Cauchy-Swarthz Inequality.

    Working the in vector space R^3 we define,
    u=(a,b,c)
    v=(b,c,a)
    Then,
    u*v<=||u||v||
    Thus,
    ab+ac+bc<=a^2+b^2+c^2
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