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More inequalities

  1. Jan 7, 2006 #1
    The numbers x and y satisfy [itex]0 < x \leq a^2, 0 < y \leq a^2, xy \geq a^2[/itex] where [itex]a \geq 1[/itex].

    By sketching suitable graphs or otherwise, show that
    [tex]x + y \geq 2a[/tex] and [itex]x \leq a^{2}y \leq a^{4}x[/itex]
    ---
    I don't know what to sketch (tried [itex]x \leq 1, y \leq 1, xy \leq 1[/itex]), so I tried algebraic methods.

    For the 1st:
    [tex]y < x+y[/tex]
    [tex]y^2 < x^2 + 2xy + y^2[/tex]
    [tex]x^2 + 2xy > 0[/tex]

    For the second one:
    [tex]x \leq a^2 \leq xy[/tex]
    [tex]a^2 \leq xy \leq a^{2}y[/tex]

    I'm really lost on this. :confused:
     
    Last edited: Jan 7, 2006
  2. jcsd
  3. Jan 7, 2006 #2
    For the first one, i get this

    [tex]xy \leq a^2y[/tex]
    [tex]xy \leq a^2x[/tex]
    [tex]2xy \leq a^2(x+y)[/tex]
    [tex]2a^2 \leq 2xy \leq a^2(x+y)[/tex]

    For the second one
    [tex]x \leq a^2[/tex]
    [tex]xy \leq a^2y[/tex]
    [tex]a^2 \leq xy \leq a^2y[/tex]
    [tex]x \leq a^2 \leq xy \leq a^2y[/tex]
    [tex]x^2 \leq a^2y[/tex]

    [tex]y \leq a^2[/tex]
    [tex]a^2y \leq a^4[/tex] since y has maximum value equal to a²
    [tex]a^2y \leq a^4 \leq a^2xy \leq a^4x[/tex]
    [tex]a^2y \leq a^4x[/tex]

    I know it is not exactly what they are asking, but it is as far as i can get from the top of my head. Others will most certainly correct me...

    marlon
     
    Last edited: Jan 7, 2006
  4. Jan 7, 2006 #3
    Thanks for your help!
     
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