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Thread: triangle ineqaulity

  1. #1
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    triangle ineqaulity

    How I prove the triangle ineqaulity by the inequalty of Cauchy-Swartz?
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    Re: triangle ineqaulity

    Quote Originally Posted by policer View Post
    How I prove the triangle ineqaulity by the inequalty of Cauchy-Swartz?
    What are you calling the inequalty of Cauchy-Swartz?
    There are several forms of inequalities that go by some form of that name.
    Please reply.
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    Re: triangle ineqaulity

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    Re: triangle ineqaulity

    Quote Originally Posted by policer View Post
    Well, in that link you are given proofs for several forms to the inequality.
    What more do you require?
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    Re: triangle ineqaulity

    I will give you an answer tomorrow I hope.
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    Re: triangle ineqaulity

    Quote Originally Posted by policer View Post
    I will give you an answer tomorrow I hope.
    You understand that we do not have a copy of the notes/textbook from which you are working?

    Here is a classic proof if one is working with absolute value as a norm.
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  7. #7
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    Re: triangle ineqaulity

    I can't solve the problem with Wikipedia.
    To solve it I need this form:
    sqrt(a^2+b^2)/(c^2+d^2) => ac + bd
    the zigzag line abc is a broken line between ab.
    ab is a straight line.
    Tell me so, how to formula it?
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  8. #8
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    Re: triangle ineqaulity

    Quote Originally Posted by policer View Post
    To solve it I need this form:
    sqrt(a^2+b^2)/(c^2+d^2) => ac + bd
    the zigzag line abc is a broken line between ab.
    ab is a straight line
    .
    I have absolutely no idea what any of that means.

    Suppose that each of $a~\&~b$ is a real number.
    $ \begin{align*}|a|\cdot |b|&=|a\cdot b| \\&\ge a \cdot b \end{align*}$ so that $2|a|\cdot |b|\ge 2a\cdot b$.

    $ \begin{align*}|a+b|^2&=(a+b)^2 \\&=a^2+2a\cdot b+b^2\\&\le |a|^2+2|a\cdot b|+b^2\\ &\le a^2+2|a|\cdot
    |b|+|b|^2\\&\le (|a|+|b|)^2 \end{align*}$
    Using the square root we get the result.
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