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Prove the Triangle Inequality

  1. Jul 5, 2005 #1
    I have proven the triangle inequality starting with ||a+b||^2 and using the Schwartz Inequality. However, the next part of the problem says:

    "Show that the Triangle Inequality is an equality if and only if |a>=alpha|b> where alpha is a real positive scalar." It must be proved in both directions.

    Any help on where to begin would be greatly appreciated.
  2. jcsd
  3. Jul 5, 2005 #2
    begin by proving one direction. complete the proof by proving the other direction. i'm not sure what |a>=alpha|b> means, is there another way to explain what that says?
  4. Jul 5, 2005 #3
    |a>=alpha|b> means the vector A equals alpha times the vector B where alpha is a real positive scalar. Does that help?

    I understand that I am "supposed" to start with one way and go the other, but what does that mean? Do I substitute a=alpha b for a and solve for ||alpha b + b|| = ||alpha b|| + ||b||? I have been playing around with the definition of ||a|| = SQRT (a a*), etc...
  5. Jul 6, 2005 #4
    in your proof of the triangle inequality make all your inequalities equalities & see what you get. at the step where you use the cauchy-schwartz inequality you see that (a,b) = |a||b|. if one vector is a multiple of the other then figure out that (a,b) = |a||b| is true. for the other way suppose that's true. then by the cauchy-schwartz inequality one vector is a multiple of the other. ( ( , ) means inner product & | | means length)
  6. Jul 6, 2005 #5


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    The triangle inequality becomes an equality when the Schawrtz inequality becomes an equality. Read through the proof of the Schwartz inequality to see when this happens.
  7. Jul 6, 2005 #6
    Start by explicitly writing out what [tex] ||~ |a \rangle + |b \rangle ~ ||^2 [/tex] is. You might start seeing where the Scwartz inequality comes into play.
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