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Yet another inequality proof

  1. Apr 19, 2007 #1
    For all real values of x and y prove that
    x^2+2xy+3y^2+2x+6y+4 >= 1

    I am trying to express the left side as a sum of sqares but always come short. Help will be appreciated.
     
    Last edited: Apr 19, 2007
  2. jcsd
  3. Apr 19, 2007 #2

    Dick

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    You are going to have trouble doing it that way. Just treat the problem as the problem of finding the minimum of the function on the LHS of your inequality. Find the point where the two partial derivatives vanish and verify the that point is a minimum and that it's >= 1.
     
  4. Apr 25, 2007 #3
    I wanna float another option. In this equation, [tex]abc+2fgh-af^2-bg^2-ch^2[/tex] is not 0. Therefore this curve represents a pair of straight lines. Geometrically this means that the pair of lines must always remain above the line y=1. The only way this is possible is if the pair of lines is parallel. If you find the equations of the two lines, then you should find that they lie above y=1.
     
  5. Apr 25, 2007 #4

    Dick

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    In what sense is a function of two variables a 'curve'? A level surface would be a curve - but which one? I don't get it.
     
  6. Apr 27, 2007 #5
    Sorry, I meant a conic section, or a pair of straight lines in the xy plane of the type [tex]ax^2+2hxy+by^2+2gx+2fy+c=0[/tex]. Here though, even if delta is not 0, it doesnt seem to be a straight line (the two constant terms are coming out to be complex). If you add 2 to both sides of the inequality, then this equation represents an ellipse with its axis rotated. So, the lowest point must lie above the line y=3.
     
    Last edited: Apr 27, 2007
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