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Another Spivak Calculus question

  1. Jul 13, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]|x + y + z| \le |x| + |y| + |z|[/tex]. Indicate when the equality holds, and prove your statement.

    2. Relevant equations
    Answer in the books says it hold only when x, y, and z are all of the same sign.


    3. The attempt at a solution
    The value on the rhs of the eq will keep getting bigger as the absolute value of x, y and z are added. It doesn't matter if x, y or z are positive or negative

    The value on the lhs will increase when the value of variable that is added is of the same sign and will decrease when the value is of the opposite sign.

    So lhs=rhs when the value of all variables are of the same sign and lhs<rhs when just one of the value are of a opposite sign.
     
  2. jcsd
  3. Jul 13, 2010 #2
    Reading between the lines, I believe your reasoning is mostly correct, although I would work harder to express it in a clearer way. The way you have it phrased is misleading. The value on the LHS doesn't need to decrease when I add a negative y to a positive x, if y is very, very negative. What matters is that it never becomes quite as large as the RHS. But why exactly?

    The inequality as given is true by the triangle inequality. So, in order to prove the answer, you need only show two things. First, if all three variables are the same sign, then there is equality. You haven't explicitly stated this in your proof. Second, if just one of the variables has a different sign than the rest, then the LHS is not equal to the RHS. You may want to use the term "without loss of generality" and specify which variable has the different sign.

    Strictly speaking, a complete answer to this question will also need to consider the case when some of the variables are equal to 0, which has no sign. This isn't difficult to do, but is something that you haven't considered.
     
  4. Jul 13, 2010 #3

    hunt_mat

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    To prove the inequality, just apply the triangle twice. It will be an equality when all signs are equal, examine a special case and things will become clear. Set x=-y and this will show that |x|+|y|=0 which can't hold for general x & y
     
  5. Jul 13, 2010 #4
    Thanks for you help, but I'm still now clear on the question.

    @ Tedjn. Yes, what I was trying to say was what you wrote in the second paragraph. By decrease I meant to say move away from the maximum value the lhs can achieve.

    @ I understand the part where there will be an equality if all the signs are the same, but I don't see why [tex]|x| + |y| = 0[/tex] if x=-y . Wouldn't that just add up to [tex]2|y|[/tex] ? I mean if y=3, the |y|=3 and |-y|=|-3|=3 ?

    So basically I should just write my proof working from the triangle inequality and considering the case where (y+z) is treated as one value

    [tex]|x + y + z| \le |x + (y + z)| \le |x| + (|y| + |z|)[/tex]

    From the triangle inequality I know that for equality to hold (y+z) must equal |y|+|z| and for that to happen y and z must be of the same sign.

    And then consider the [tex] |x + (y + z)| \le |x| + (|y| + |z|)[/tex]. So x has to be the same sign as (y+z) for equality to hold.

    And also from what I've worked out about the triangle inequality, if the sign is different, lhs will be less than rhs.

    Is that how it goes?

    Thank you for kindly helping me out.
     
  6. Jul 13, 2010 #5

    Hurkyl

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    Gold Member

    Yes, it would, which is why it's generally problematic for that to equal 0. Why is it zero? Well, that's obvious -- if you don't forget the other equation you're assuming.
     
  7. Jul 13, 2010 #6
    Sorry, I really can't figure out what's the other equation I'm assuming that will make the sum 0. Please help.
     
  8. Jul 14, 2010 #7

    hunt_mat

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    Think of it like this, take x=-1,y=1 and z=1, then if it were equality then 1=3 which is not the case and equality would hold. Why are you so hung on on this any way? You know from the normal triangle inequality that [tex]|x+y|\leq |x|+|y|[/tex] with equality holding true if and only if x & y have the same sign right? If you can prove this for in usual triangle inequality then you're done, as all you do is apply the triangle equality twice and then at every stage you say equality holds for the same sign.
     
  9. Jul 14, 2010 #8

    hunt_mat

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    Note that in general [tex]|x|+|y|\geq 0[/tex] with equality holding if and only if x=y=0.

    Mat
     
  10. Jul 14, 2010 #9
    Thanks, hung_mat, I finally understand what you were trying to say.
     
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