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Triangle inequality

  1. Jul 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Show for nonnegative x,y,z that

    [tex](x+y+z) \sqrt{2} \leq \sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}[/tex]

    My book says the answer comes from apply the triangle inequality to (x+y+z,x+y+z)=(x,y)+(y,z)+(x,z). I don't see what they mean by that at all. HOW do you apply the triangle inequality to that?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jul 20, 2008 #2


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    The standard form of the triangle inequalilty is [itex]d(x,y)\le d(x,z)+ d(y,z)[/itex]. But it is also true that [itex]d(x,z)\le d(z, u)+ d(u, z)[/itex] so, putting those together, you can say that [itex]d(x,y)\le d(x,u)+ d(u,z)+ d(z,y)[/itex]. Do you see to apply that?
  4. Jul 20, 2008 #3


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    Take the Euclidian norm of both sides of the equation. For instance, the LHS becomes

  5. Jul 20, 2008 #4
    I see. Thanks.
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