1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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...

    Last edited: Jan 7, 2006
  4. Jan 7, 2006 #3
    Thanks for your help!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook