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Math Help - Vector Proof

  1. #1
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    Vector Proof

    Hi all.

    I am stuck at this vector proof question and need help.

    Suppose vectors a,b,c satisfy ||a|| > (1/2) + sqrt((1/4)+||c||)

    and ||b|| = (||a||)^2

    Prove that ||b-c|| > ||a||

    Someone please help, thank you.
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  2. #2
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    Re: Vector Proof

    You have \displaystyle \begin{align*} |\mathbf{a}| > \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} \end{align*} and \displaystyle \begin{align*} |\mathbf{b}| = |\mathbf{a}|^2 \end{align*}.

    From the first inequality

    \displaystyle \begin{align*} |\mathbf{a}|^2 &> \left( \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} \right)^2 \\  |\mathbf{a}|^2 &> \frac{1}{4} + \sqrt{\frac{1}{4} + |\mathbf{c}|} + \frac{1}{4} + |\mathbf{c}| \\ |\mathbf{a}|^2 &> \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} + |\mathbf{c}| \\ |\mathbf{b}| &> \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} + |\mathbf{c}| \\ |\mathbf{b}| - |\mathbf{c} | &> \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} \\ |\mathbf{b} - \mathbf{c}| &> \frac{1}{2} + \sqrt{\frac{1}{4} + |\mathbf{c}|} \end{align*}

    I'm afraid I don't know how to continue...
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  3. #3
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    Re: Vector Proof

    We have
    |\mathbf{a}| - 1/2 > \sqrt{1/4 + |\mathbf{c}|}
    so (squaring),

    |\mathbf{a}|^2 - |\mathbf{a}| + 1/4 > 1/4 + |\mathbf{c}|

    hence
    |\mathbf{a}|^2   - |\mathbf{c}| > |\mathbf{a}|

    substituting |\mathbf{b}| = |\mathbf{a}|^2,

    |\mathbf{b}|   - |\mathbf{c}| > |\mathbf{a}|

    But by the triangle inequality,
    |\mathbf{b} - \mathbf{c}| + |\mathbf{c}| \ge |\mathbf{b} - \mathbf{c} +  \mathbf{c}| = |\mathbf{b}| ,
    so
    |\mathbf{b} - \mathbf{c}| \ge |\mathbf{b}| - |\mathbf{c}|

    Consequently,
    |\mathbf{b} - \mathbf{c}| \ge |\mathbf{b}| - |\mathbf{c}| > |\mathbf{a}|
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