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Math Help - Proof of triangle inequality.

  1. #1
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    Proof of triangle inequality.

    Prove:
    The sum of the lengths of any two sides of a triangle is greater than the length of the third side. This is known as the triangle inequality.

    I need some help with this proof.

    Thanks in advance for any help.
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  2. #2
    Super Member Aryth's Avatar
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    This may not be the best way to do it, but it's my favorite method.

    Since we are trying to prove that:

    |a + b| \leq |a| + |b|

    It suffices to show that:

    |a+b|^2 \leq (|a| + |b|)^2

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

    = a^2 + 2ab + b^2

    = |a|^2 + 2ab + |b|^2

    \leq |a|^2 + 2|a||b| + |b|^2

    \leq (|a| + |b|)^2


    The one that everyone else might use is:

    |a+b| \leq |a| + |b|

    Notice that -|a| \leq a \leq |a| and that -|b| \leq b \leq |b|.

    When we add those two statements together, we get:

    -(|a| + |b|) \leq a+b \leq (|a| + |b|)

    by the absolute value property, you get:

    |a+b| \leq |a| + |b|
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  3. #3
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    Krizalid's Avatar
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    Aryth, you missunderstood the question, read it again. The original one doesn't involve the absolute value.
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  4. #4
    Super Member Aryth's Avatar
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    Ah, well, the triangle inequality I was using was:

    The triangle inequality states that for any triangle, the length of a given side must be less than or equal to the sum of the other two sides but greater than or equal to the difference between the two sides.

    In a normed vector space V, the triangle inequality is:

    ||x+y|| \leq ||x|| + ||y||, \ \forall \ x,y \in V

    Since the real line is a normed vector space, then the triangle inequality for real numbers x and y is:

    |x+y| \leq |x| + |y|

    So... What triangle inequality should I be proving?
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