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Math Help - Absolute value inequality (and a square root, sadly)

  1. #1
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    Question Absolute value inequality (and a square root, sadly)

    Hi guys,

    Given: \sqrt{|1-x|} \geq x
    Expanding absolute value: x > 1 \wedge {\color{Red} \sqrt{x-1} \geq x} \vee x \leq 1 \wedge \sqrt{1-x} \geq x
    Note: the inequality marked in red is never true. I thinks this OK because the given is also never true for x > 1.

    Questions: Have I expanded the absolute value sign correctly? If not: any hints?

    Many thanks
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Absolute value inequality (and a square root, sadly)

    Hi edg85!

    Quote Originally Posted by edg85 View Post
    Hi guys,

    Given: \sqrt{|1-x|} \geq x
    Expanding absolute value: x > 1 \wedge {\color{Red} \sqrt{x-1} \geq x} \vee x \leq 1 \wedge \sqrt{1-x} \geq x
    Note: the inequality marked in red is never true. I thinks this OK because the given is also never true for x > 1.

    Questions: Have I expanded the absolute value sign correctly? If not: any hints?

    Many thanks
    Yes.
    All is well.
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  3. #3
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    Re: Absolute value inequality (and a square root, sadly)

    Quote Originally Posted by edg85 View Post
    Given: \sqrt{|1-x|} \geq x

    STOP, step back and think. This is a beautiful thought question.

    Is it true for any x\le 0~?

    For what x>0 is it true?

    When is x>0\text{ and }|x-1|>x^2\text{ true ?}

    Did you plot \sqrt{|1-x|}-x~?
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  4. #4
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    Re: Absolute value inequality (and a square root, sadly)

    Hi, thanks for the replies. @Plato yes i have a plot, handdrawn actually. Something wrong? Have i expanded the absolute value correctly in my first post? If so i will post the rest of my solution asap.
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    Re: Absolute value inequality (and a square root, sadly)

    Quote Originally Posted by edg85 View Post
    Hi, thanks for the replies. @Plato yes i have a plot, handdrawn actually. Something wrong? Have i expanded the absolute value correctly in my first post? If so i will post the rest of my solution asap.
    Can you show why
    x\le \frac{-1+\sqrt5}{2} is a solution?
    Last edited by Plato; February 26th 2013 at 06:27 PM.
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  6. #6
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    Re: Absolute value inequality (and a square root, sadly)

    Quote Originally Posted by Plato View Post
    Can you show why
    x\le \frac{-1+\sqrt5}{2} is a solution?
    Sure! Let's go:

    Given: \sqrt{|1-x|} \geq x
    Expanding absolute value:
    x > 1 \wedge {\color{Red} \sqrt{x-1} \geq x} \vee x \leq 1 \wedge \sqrt{1-x} \geq x
    Squaring:
    x > 1 \wedge {\color{Red} x-1 \geq x^2} \vee x \leq 1 \wedge 1-x \geq x^2
    Sorting:
    x > 1 \wedge {\color{Red} x^2-x+1 \leq 0} \vee x \leq 1 \wedge x^2+x-1 \leq  0
    Solving quad. ineq.:
    x > 1 \wedge \text{{\color{Red} no x}} \vee x \leq 1 \wedge \frac{-\sqrt{5}-1}{2} \leq x \leq \frac{\sqrt{5}-1}{2}
    Combining:
    x \leq \frac{\sqrt{5}-1}{2}
    Note: the inequality marked in red is never true. I thinks this OK because the given is also never true for x > 1.
    Note 2: i don't know any symbol/notation for "no x".

    Conclusion:
    \sqrt{|1-x|} \geq x if \left \{ x \in \mathbb{R}\mid x \leq \frac{\sqrt{5}-1}{2} \right \}

    Does this make sense at all?
    Last edited by edg85; February 27th 2013 at 01:36 AM.
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