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A Triangle inequality question

  1. Jun 21, 2016 #1
    In the derivation of triangle inequality [tex]|(x,y)| \leq ||x|| ||y||[/tex] one use some ##z=x-ty## where ##t## is real number. And then from ##(z,z) \geq 0## one gets quadratic inequality
    [tex]||x||^2+||y||^2t^2-2tRe(x,y) \geq 0[/tex]
    And from here they said that discriminant of quadratic equation
    [tex]D=4(Re(x,y))^2-4 ||y||^2|x||^2 \leq 0 [/tex]
    Could you explain me why ##<## sign in discriminant relation? When discriminant is less then zero solutions are complex conjugate numbers. But I still do not understand the discussed inequality. What about for example in case
    [tex]||x||^2+||y||^2t^2-2tRe(x,y) \leq 0[/tex]?
     
  2. jcsd
  3. Jun 21, 2016 #2

    blue_leaf77

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    The quadratic equation
    $$
    ||x||^2+||y||^2t^2-2tRe(x,y) \geq 0
    $$
    must always be zero or positive for all real ##t##. This means the function ##f(t)=||x||^2+||y||^2t^2-2tRe(x,y)## must always be above the ##t## axis, which further implies that it can have at most one solution (which occurs when the minimum of this function touches the ##t## axis): one solution or no solution at all. In terms of the discrimant, the discrimant of ##f(t)## is either zero or negative.
     
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