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Math Help - Another proof

  1. #1
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    Another proof

    I have to Prove (a+b)^2 + (b+c)^2 + (c+a)^2 >= 4(ab + bc + ac)

    I multiplied out both sides and then move everything to the left, but I don't know what I should do after that.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Nichelle14
    I have to Prove (a+b)^2 + (b+c)^2 + (c+a)^2 >= 4(ab + bc + ac)

    I multiplied out both sides and then move everything to the left, but I don't know what I should do after that.
    (a+b)^2 + (b+c)^2 + (c+a)^2
    =2(a^2 + b^2 + c^2)+ 2(ab + bc + ca)
    =(a^2 + b^2)+(b^2 + c^2)+(a^2 + c^2)+ 2(ab + bc + ca)
    \geq4(ab + bc + ca)

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  3. #3
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    Quote Originally Posted by Nichelle14
    I have to Prove (a+b)^2 + (b+c)^2 + (c+a)^2 >= 4(ab + bc + ac)

    I multiplied out both sides and then move everything to the left, but I don't know what I should do after that.
    It is necessary and suffienct for,
    a^2+b^2+c^2\geq ab+ac+bc
    ----
    If you play your cards right this is the "Cauchy-Swarchtz Inequality".
    Let,
    \bold{v}=(a,b,c)
    \bold{u}=(b,c,a)
    Then,
    ||\bold{u}||\cdot ||\bold{v}||\geq |\bold{u}\cdot \bold{v} |
    Now,
    <br />
||\bold{u}||\cdot ||\bold{v}||=a^2+b^2+c^2
    And,
    |\bold{u}\cdot \bold{v} |=|ab+bc+ac|
    Thus,
    a^2+b^2+c^2\geq |ab+bc+ac|\geq ab+bc+ac
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  4. #4
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    Hello, Nichelle14!


    Prove that: . (a+b)^2 + (b+c)^2 + (c+a)^2\:\geq\:4(ab + bc + ac)

    I multiplied out both sides and then move everything to the left. . . . Not recommended

    Multiply out the left side and we have:
    . . (a+b)^2 + (b+c)^2 + (c+a)^2\:= \:2(a^2 + b^2+c^2) + 4(ab + bc + ac) LHS


    Since a,\,b,\,c are real numbers: . a^2 + b^2 + c^2 \:\geq \:0

    Multiply by 2:\;\;2(a^2 + b^2 + c^2) \:\geq \:0


    Add 4(ab + bc + ac) to both sides:

    . . \underbrace{2(a^2 + b^2 + c^2) + 4(ab + bc + ac)} \:\geq\:4(ab + bc + ac)
    . . . . . . . . This is LHS

    Therefore: . (a+b)^2 + (b+c)^2 + (c+a)^2\;\geq\;4(ab + bc + ac)

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