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Inequality problem

  1. Jul 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##x^2+y^2+xy+1 \geq a(x+y)## for all ##x,y \in R##. Find the possible integer(s) in the range of ##a##.


    2. Relevant equations



    3. The attempt at a solution
    I can rewrite this into ##(x+y)^2-xy+1 \geq a(x+y) \Rightarrow (x+y)(x+y-a)+1-xy \geq 0## but I don't think that this would help.
     
  2. jcsd
  3. Jul 11, 2013 #2
    You can treat this as a quadratic inequality with a variable [itex]x[/itex]:

    [itex]x^2+(y-a)x+y^2-ay+1\ge 0[/itex]

    and determine when this parabola takes only non-negative values.
     
  4. Jul 11, 2013 #3

    verty

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    I don't have any great advice but the method of Lagrange multipliers (from calculus) shows that the = case happens when x = y. Probably one must by some argument show that the strongest constraining effect happens when x = y. Or perhaps not, but I don't have any better advice.
     
    Last edited: Jul 11, 2013
  5. Jul 11, 2013 #4

    verty

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    I've got it. You must rotate the coordinate axes. Rotate them 45 degrees counter-clockwise, this will make the right side depend on a single variable. I was looking for a way to make this problem easier.

    ##(x,y) → (\frac{x'}{\sqrt{2}} - \frac{y'}{\sqrt{2}}, \frac{x'}{\sqrt{2}} + \frac{y'}{\sqrt{2}})##
     
  6. Jul 11, 2013 #5

    haruspex

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    It doesn't need to be a pure rotation, so you can forget the √2s. Any invertible substitution is valid.
     
  7. Jul 12, 2013 #6
    You have already found one equation;
    ##(x+y)^2-xy+1 \geq a(x+y)##
    This can again be rewritten as ##(x+1)^2+(y+1)^2-2(x+y)+xy-1\geq a(x+y)##

    Add these two equations and with little manupulation you get,

    ##(x+1)^2+(y+1)^2+(x+y)^2 \geq 2(a+1)(x+y)##

    Then with some more manupulation you get,

    ##-(x+y)(x+y-2a-2) \geq (x+1)^2+(y+1)^2## Divide by 2..

    ##(-(x+y)(x+y-2a-2))/2 \geq ((x+1)^2+(y+1)^2)/2##


    here ##(x+1)^2## and ##(y+1)^2## are both positive. So, their arithmetic mean is ≥ geometric mean
    which implies,

    ##(-(x+y)(x+y-2a-2))/2 \geq sqrt(x+1)(y+1)##

    I think we have got nearer to the answer. Hope this helps
     
    Last edited: Jul 12, 2013
  8. Jul 12, 2013 #7

    verty

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    Right, I didn't see that. So any invertible substitution that takes x+y to a single-variable alternative will work to make the problem solvable without calculus or advanced techniques. The basic idea is to let a depend on x only, so we can set y to 0 without affecting anything.
     
  9. Jul 13, 2013 #8
    Sorry for such a late reply.

    The discriminant should be less than zero i.e.
    [tex](y-a)^2-4(y^2-ay+1)<0[/tex]
    What am I supposed to do with this? Work on the discriminant of the resulting quadratic again? :confused:

    Is it okay to add them? :uhh:
     
  10. Jul 13, 2013 #9
    The discriminant should be less or equal zero as the value must be non-negative, not positive. And yes, work on the discriminant again.
     
  11. Jul 13, 2013 #10
    Yes you can add them.
    If a<b and c<d
    then a+b<c+d..
     
  12. Jul 14, 2013 #11
    Thank you! That is a nice method. :smile:
     
  13. Jul 14, 2013 #12
    You are welcome :smile:
     
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