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Math Help - Inequality Proof

  1. #1
    Junior Member
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    Inequality Proof

    Hi I am working on my homework for Discrete Math and I don't know how to do this proof. It says, "Use proof by cases that |x+y|≤ |x| + |y| for all real numbers x and y." I know what proof by cases is but I don't know where to start.

    Ok....I have started doing that and this is what I have so far:

    Case 1: x + y>0
    |x+y|>0 → |x+y| = x + y.

    x + y cannot be > |x| + |y|.
    ∴ x + y ≤ |x| + |y|


    Case 2: x + y = 0
    |x + y| = 0 → x=x & y= -x

    ∴ x + (-x) ≤ |x| + |y|

    Case 3: x + y < 0
    |x + y| < 0 → x + y = -(x + y).

    I have gotten this far, but I donít know where to go from here. I know it makes sense in my head, but I donít know how to put it into words.
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  2. #2
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    Here is another way.
    \begin{gathered}0 \leqslant x \Rightarrow \quad x = \left| x \right| \Rightarrow \quad x \leqslant \left| x \right| \hfill \\ x < 0 \Rightarrow \quad x < 0 <  - x = \left| x \right| \hfill \\  \left( {\left| {x + y} \right|} \right)^2 =\left( {x + y} \right)^2  = x^2  + 2xy + y^2  \leqslant \left| x \right|^2  + 2\left| x \right|\left| y \right| + \left| y \right|^2 =\left( {\left| x \right| + \left| y \right|} \right)^2  \hfill \\\end{gathered}
    Now take the square root.
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  3. #3
    Junior Member
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    i understand the first two cases, but I don't understand the reason why you squared x + y in the last case.
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