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Math Help - Using triangle inequality to prove inequalities

  1. #1
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    Using triangle inequality to prove inequalities

    I have been given the triangle inequality:

    ||x| - |y|| <= |x + y| <= |x| + |y|

    where |x| is the absolute value of x.

    Now I'm supposed to prove the following:

    |3x + 2| + |3x - 2| >= 4 for all real x.

    Now, I tried this:

    |3x + 2| + |3x - 2| >= |3x + 2 + 3x - 2| = 6|x|.

    This does not solve the problem. How do I do this?
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  2. #2
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    |3x + 2| = \begin{cases} 3x + 2 & x \geq -  \frac{2}{3}  \\  2 - 3x & x < - \frac{2}{3}\end{cases}

    We can see that its minimum occurs at x = -\frac{2}{3}, i.e |3x + 2| \geq 2

    We also find the same for |3x - 2| \geq 2

    Conclusion follows
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  3. #3
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    o_O: Sorry but that does not answer my question. I am supposed to use the triangle inequality to prove this.
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  4. #4
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    Haha look at this ^^


    |3x-2|=|2-3x| by the property of absolute value, that is |a|=|-a|


    Now consider |3x+2|+|2-3x|
    this is in the form |x|+|y|, which is \geq |x+y| by triangle inequality.

    Hence |3x+2|+|2-3x| \geq |3x+2+2-3x|=4

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  5. #5
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    Hahah, wow. How did I miss that? Thanks!
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