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Homework Help: Maybe triangle inequalities theorem

  1. Aug 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove: If a2+b2=1, and c2+d2=1, then ac+bd ≤1

    2. Relevant equations

    Maybe triangle inequalities theorem.

    3. The attempt at a solution
  2. jcsd
  3. Aug 2, 2012 #2


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    Re: Inequalities

    For this question to make sense, I think a,b,c and d must be Real numbers. Is this given?

    Hint: What is (a+c)2+(b+d)2?
  4. Aug 2, 2012 #3
    Re: Inequalities

    At the beginning of the chapter, "An inequalities is a statement that one(Real)number is greater than or less than another".
    Is there inequalites in complex number?

    equal zero if both factors=0, greater than zero if one or both factor not equal to zero.
    a2+c2+2ac +b2+d2+2bd≥0
    ac+bd≤-1 ???

    So it should be
    Equal to zero if a=c and b=d, greater than zero if one or both of components of the factors not equal.
    a2+c2-2ac +b2+d2-2bd≥0
    Last edited: Aug 2, 2012
  5. Aug 2, 2012 #4


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    Re: Inequalities

    One can extend the definition of an inequality to deal with complex numbers (by comparing their real and imaginary parts), but this isn't nessecary in this case since if [itex]a[/itex] and [itex]c[/itex] are both purely imaginary, their product will still be real (but the inequality that you are supposed to prove may not be satisfied).

    Anyways, if you are studying real numbers then it is probably safe to assume all 4 numbers (a,b,c & d) are real.

    This is true, but not all that useful here. instead of looking at the lower limit of (a+c)2+(b+d)2, use the triangle inequality (twice) to look at its upper limit.

    Keep in mind, that if x2+y2=1 (for real-valued x & y), then |x|^2=x^2=1-y^2≤1 :wink:

    This method also works and is even simpler than mine!:approve:
  6. Aug 3, 2012 #5


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    Re: Inequalities

    But (not relevant to this question but just so that other readers do not get the wrong idea) that extension of inequality does not make the complex numbers an ordered field. In particular, if x< y and 0< z, it is NOT necessarily true that xz< yz.

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