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Triangle Inequality Question

  1. Jul 14, 2013 #1
    I'm beginning to read Spivak's Calculus 3ed, and everything is smooth until I reach page 12.

    QNaq3XL.png

    My question is marked, between line 2 and 3. Why there's such sign change suddenly? In fact I tried with simple line 4 case and it's not in fact equal. I'm assuming that a and b is valid for all integer case whether they are negative or not.

    Then I read this:
    http://math.ucsd.edu/~wgarner/math4c/derivations/other/triangleinequal.htm [Broken]
    (Please see to the Alternative Proof of the Triangle Inequality section)

    It clearly contradicts what Spivak's book said in line 3. Then I think, whether he intends to do the case where a and b are both positive, but then the question arises why there's larger than sign in line 2 if that's the case.

    Thanks
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 14, 2013 #2

    jbunniii

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    To go from line 1 to line 2, note that for any real number ##x##, we have ##x \leq |x|##. In particular, for ##x = ab##, we have ##ab \leq |ab|##. But of course ##|ab| = |a| \cdot |b|##, so the inequality becomes ##ab \leq |a| \cdot |b|##. We may now multiply this by 2, and add ##a^2 + b^2## to both sides to obtain
    $$a^2 + 2ab + b^2 \leq a^2 + 2|a| \cdot |b| + b^2$$
    To go from line 2 to 3, we simply recognize that since ##a## and ##b## are real, we have ##a^2 = |a|^2## and ##b^2 = |b|^2##. Therefore the right hand side of the above inequality is equal to ##|a|^2 + 2 |a| \cdot |b| + |b|^2##.

    Combining all of the above, we conclude that
    $$\begin{align}a^2 + 2ab + b^2 &\leq a^2 + 2|a| \cdot |b| + b^2 \\
    &= |a|^2 + 2|a| \cdot |b| + |b|^2\end{align}$$
    The proof at your link carries out the same steps, but in a different order, which is why the ##\leq## appears later.

    Note that the ##=## sign in the above chain does NOT mean that all of the expressions are equal. Only the expressions immediately before and after the ##=## sign are equal. Since there is at least one ##\leq## in the chain, the correct conclusion is
    $$(|a|+|b|)^2 \leq |a|^2 + 2|a| \cdot |b| + |b|^2$$
    not
    $$(|a|+|b|)^2 = |a|^2 + 2|a| \cdot |b| + |b|^2$$
     
    Last edited: Jul 14, 2013
  4. Jul 14, 2013 #3
    Ah ok! I get what you're explaining. It seems clearer now. What I didn't get is the fact that the sign in line 2 is 'nested' to that of line 3. Not to the first original expression on the far left. You left out the factor 2 also but I assumed we can just multiply that in and adds ##a^2## and ##b^2##, in the beginning of your explanation.

    (I was quite sleepy perhaps, sorry)

    Do you mean the last part as this?
    $$(|a+b|)^2 \neq |a|^2 + 2|a| \cdot |b| + |b|^2$$

    Because I think this is true. (I added the missing 2, because without it even the first inequality wouldn't hold)
    $$(|a|+|b|)^2 = |a|^2 + 2|a| \cdot |b| + |b|^2$$

    and thus
    $$(|a+b|)^2 \leq |a|^2 + 2|a| \cdot |b| + |b|^2 = (|a|+|b|)^2$$

    Thanks for your help!
     
    Last edited: Jul 14, 2013
  5. Jul 14, 2013 #4

    jbunniii

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    Yes, sorry, I left the 2 out of both expressions. I'll edit my post now to avoid confusion.
     
  6. Jul 14, 2013 #5

    jbunniii

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    Yes, that's right. In general, something of the form ##X \leq Y = Z## means "##X \leq Y## and ##Y = Z##", from which it follows that ##X \leq Z##.
     
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