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Math Help - [SOLVED] Absolute values

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    [SOLVED] Absolute values

    My own problem! Enjoy.

    Show that for a,b,c \in \mathbb{R}, one has

    |a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |
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    Quote Originally Posted by Bruno J. View Post
    My own problem! Enjoy.

    Show that for a,b,c \in \mathbb{R}, one has

    |a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |

    We need to show that it is an identity which holds for all real numbers  a ,b ,c ?
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Yes!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Yes!
    Is there a clever way to do this or just brute stuff? If you created it you probably worked backwards from something and so have an "ideal" solution which retraces your steps. I would guess the easiest way to do this would be to have that f(a,b,c) be the LHS and then we are trying to show that f(a,b,c)=f(a,c,b) and to start squaring stuff.
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    Quote Originally Posted by Drexel28 View Post
    Is there a clever way to do this or just brute stuff? If you created it you probably worked backwards from something and so have an "ideal" solution which retraces your steps. I would guess the easiest way to do this would be to have that f(a,b,c) be the LHS and then we are trying to show that f(a,b,c)=f(a,c,b) and to start squaring stuff.

    Clever way ? Is it useful to draw a geometric figure ? Haha , sounds quite silly .
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Is there a clever way to do this or just brute stuff? If you created it you probably worked backwards from something and so have an "ideal" solution which retraces your steps. I would guess the easiest way to do this would be to have that f(a,b,c) be the LHS and then we are trying to show that f(a,b,c)=f(a,c,b) and to start squaring stuff.
    There is a very clever way! A small hint would give it all away... You don't even really have to do any calculations, even less draw a diagram!
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Bruno J. View Post
    My own problem! Enjoy.

    Show that for a,b,c \in \mathbb{R}, one has

    |a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |
    This approach is basically brute force...

    There are 3! ways to permute {a,b,c}, leading to 6 cases:

    case a \leq b \leq c

    |a-b|+c+\Big | a+b+|a-b|-2c\Big | =?\ |a-c|+b+\Big |  a+c+|a-c|-2b\Big |

    b-a+c+ | a+b+b-a-2c | =?\ c-a+b+|  a+c+c-a-2b |

    b-a+c+ | 2b-2c | =?\ c-a+b+| 2c-2b |

    b-a+c+ 2(c-b) = c-a+b+2(c-b)

    I won't list the other cases because they all work out similarly, unless I'm missing something.
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    Quote Originally Posted by Bruno J. View Post
    My own problem! Enjoy.

    Show that for a,b,c \in \mathbb{R}, one has

    |a-b|+c+\Big | a+b+|a-b|-2c\Big | = |a-c|+b+\Big | a+c+|a-c|-2b\Big |
    We have \max\{a,b\} = \frac{a+b+|a-b|}{2}. Thus,

    \max\{a,b,c\} = \max\{a,\max\{b,c\}\} = \frac{1}{4}(2a+b+c+|b-c|+\Big|b+c+|b-c|-2a\Big|)

    We can switch a, b and c in the above identity and still get \max\{a,b,c\}. So:

    \text{RHS when } a \leftrightarrow c = \text{RHS when } a \leftrightarrow b

    <br />
\frac{1}{4}(2c+a+b+|a-b|+\Big|a+b+|a-b|-2c\Big|) =<br />
\frac{1}{4}(2b+a+c+|a-c|+\Big|a+c+|a-c|-2b\Big|)<br />

    Multiplying by 4 and subtracting a+b+c from both sides yields the result.
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    We have \max\{a,b\} = \frac{a+b+|a-b|}{2}. Thus,

    \max\{a,b,c\} = \max\{a,\max\{b,c\}\} = \frac{1}{4}(2a+b+c+|b-c|+\Big|b+c+|b-c|-2a\Big|)

    We can switch a, b and c in the above identity and still get \max\{a,b,c\}. So:

    \text{RHS when } a \leftrightarrow c = \text{RHS when } a \leftrightarrow b

    <br />
\frac{1}{4}(2c+a+b+|a-b|+\Big|a+b+|a-b|-2c\Big|) =<br />
\frac{1}{4}(2b+a+c+|a-c|+\Big|a+c+|a-c|-2b\Big|)<br />

    Multiplying by 4 and subtracting a+b+c from both sides yields the result.
    Good job!
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