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Spivak inequality

  1. May 21, 2013 #1

    Find all numbers [itex]x[/itex] for which [itex]\frac{1}{x}+\frac{1}{1-x}>0[/itex].


    If [itex]\frac{1}{x}+\frac{1}{1-x}>0[/itex],

    then [itex]\frac{1-x}{x(1-x)}+\frac{x}{x(1-x)}>0[/itex];

    hence [itex]\frac{1}{x(1-x)}>0[/itex].

    Now we note that

    [itex]\frac{1}{x(1-x)} \rightarrow ∞[/itex] as [itex]x \rightarrow 0[/itex]

    and [itex]\frac{1}{x(1-x)} \rightarrow 0[/itex] as [itex]x \rightarrow 1[/itex].

    Thus, [itex]0<x<1[/itex].


    Not quite sure if that's the sort of solution Spivak is looking for in Ch.1.
  2. jcsd
  3. May 21, 2013 #2
    A non-zero number and its reciprocal will always have the same sign so [itex] \frac{1}{x(1-x)} [/itex] will be positive where [itex] x(1-x) [/itex] is
  4. May 21, 2013 #3
    Ah, I see. Don't know how I didn't see that.
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