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

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Homework Help Overview

The discussion revolves around proving the inequality x^y > y^x for the case where y > x ≥ e. Participants are exploring various mathematical approaches to establish this relationship.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use logarithmic manipulation and the Hessian determinant to analyze the function g(x,y) = y*ln(x) - x*ln(y). They question the validity of using the Hessian approach and seek alternative methods, including induction.
  • Another participant suggests fixing x and analyzing g(y) to demonstrate properties of the function, including its derivatives.
  • There is a suggestion to simplify the problem by substituting y with kx, prompting further exploration of this approach.

Discussion Status

The discussion is ongoing, with participants providing different perspectives on the problem. Some guidance has been offered regarding the use of derivatives and function behavior, but no consensus has been reached on the best method to prove the inequality.

Contextual Notes

Participants are navigating the constraints of the problem, particularly the conditions y > x ≥ e, and are questioning the appropriateness of various mathematical techniques, including the Hessian determinant.

viliperi
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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.

Homework Statement



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

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?
 
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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.


viliperi said:
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.

Homework Statement



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

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?
 
welcome to pf!

hello viliperi! welcome to pf! :smile:

or you could just put y = kx, and simplify :wink:
 
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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