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Math Help - What do I do next

  1. #1
    Junior Member
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    What do I do next

    I am trying to prove that for any real numbers a,b,c the follwing inequality holds:

    (a^2 + b^2 + c^2)/3 >= ((a+b+c)/3)^2

    I multiplied out the square on the RHS and simplified.
    Then got rid of the fraction on each side.

    But what do I do next?
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    I am trying to prove that for any real numbers a,b,c the follwing inequality holds:

    (a^2 + b^2 + c^2)/3 >= ((a+b+c)/3)^2

    I multiplied out the square on the RHS and simplified.
    Then got rid of the fraction on each side.

    But what do I do next?
    I am using a different method here.
    LHS= \frac{a^2 + b^2 + c^2}{3}=\frac{3(a^2 + b^2 + c^2)}{9}
    =\frac{a^2 + b^2 + c^2}{9}+\frac{2(a^2 + b^2 + c^2)}{9}
    \geq\frac{a^2 + b^2 + c^2}{9}+\frac{2(ab+bc+ca)}{9}
    \geq\frac{(a + b + c)^2}{9}

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    Malay
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  3. #3
    Super Member malaygoel's Avatar
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    Smile

    Quote Originally Posted by Nichelle14
    I am trying to prove that for any real numbers a,b,c the follwing inequality holds:

    (a^2 + b^2 + c^2)/3 >= ((a+b+c)/3)^2

    I multiplied out the square on the RHS and simplified.
    Then got rid of the fraction on each side.

    But what do I do next?
    I think you obtained the equation
    2(a^2+b^2+c^2+ab+bc+ca)\geq0
    Do like this
    (a^2+b^2+2ab)+(b^2+c^2+2bc)+(c^2+a^2+2ac)\geq0

    KeepSmiling
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  4. #4
    Junior Member
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    that's clever.
    I did get down to

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

    and figured it out. I like your way too!
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  5. #5
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    If x denotes a sequence of numbers, and AM(x) its arithmetic mean, you're being asked to prove that AM(f(x)) >= f(AM(x)) where f is squaring and f(x) means the sequence of values of f applied to the elements of x. You could prove this using only the fact that f is convex (f' and f'' both positive).
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