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Prove that x^y > y^x when y>x>e

  1. Mar 1, 2012 #1
    Hello everybody! I am new here and probably going to visit here pretty often in the future. Sometimes I do not get very clear answer from the lecturers so I hope to get more sensible answers from PF.

    1. The problem statement, all variables and given/known data

    I need to show that x^y > y^x whenever y > x ≥ e.

    3. The attempt at a solution


    At first I start by multiplying by ln(): y*ln(x) > x*ln(y)

    Then I take g(x,y) such that g(x,y) = y*ln(x) - x*ln(y) and 2nd order derivatives and show det(H(g)) < 0 whenever y > x ≥ e

    My purpose with this is to show that there are no real local or global critical points in g(x,y) when y > x >= e, and conclude that g(x) = x^y - y^x diverges. I was not told why I can not use Hessian determinant to make this conclusion.

    Why I can not use Hessian approach to prove this? What would be the correct way to do it? Can I prove it with induction?
     
  2. jcsd
  3. Mar 2, 2012 #2
    You may let x be fixed and let g(y)=y*ln(x)-x*ln(y).

    The first step, you to show that g(x)=0, (say, let y=x)

    2nd, g'(y)=ln(x)-x/y, and we have g'(x)=ln(x)-1>0 (since x>3);

    3rd, g''(y)=x/y^2>0, so g'(y) is monotonically increasing in y>x>e, and therefore, g'(y)>0 for y>x>e.

    and complete the proof.

    Good luck.


     
  4. Mar 2, 2012 #3

    tiny-tim

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    welcome to pf!

    hello viliperi! welcome to pf! :smile:

    or you could just put y = kx, and simplify :wink:
     
  5. Mar 2, 2012 #4
    Thanks for that! I'll try that.

    But you agree with my intuition that it is suitable to utilise Hessian approach to this exercise? Is there any other way to conclude that it is always true with given conditions?
     
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