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Let $x>y>0$. Prove that $$x^{y^x}>y^{x^y}.$$ My attempts:

  1. Let $1>x>y>0$.

In this case it's enough to prove that $$y^x<x^y$$ or $$x\ln\frac{1}{y}>y\ln\frac{1}{x},$$ which is obvious;

  1. $x\geq1>y>0$.

In this case our inequality is obviously true;

  1. $e\geq x>y>1.$.

We need to prove that $f(x)\geq0,$ where $$f(x)=x\ln{y}-y\ln{x}+\ln\ln{x}-\ln\ln{y}.$$ Now, $$f'(x)=\ln{y}-\frac{y}{x}+\frac{1}{x\ln{x}}.$$ Let $h(y)=\ln{y}-\frac{y}{x}+\frac{1}{x\ln{x}}.$

Thus, $$h'(y)=\frac{1}{y}-\frac{1}{x}>0,$$ which says $$h(y)>h(1)=-\frac{1}{x}+\frac{1}{x\ln{x}}=\frac{1-\ln{x}}{x\ln{x}}\geq0.$$ Id est, $f$ increases and $$f(x)>f(y)=0;$$ 4. $x>y\geq e$.

Since $$\left(\frac{\ln{x}}{x}\right)'=\frac{1-\ln{x}}{x^2}\leq0$$ for all $x\geq e,$ we obtain: $$f(x)=xy\left(\frac{\ln{y}}{y}-\frac{\ln{x}}{x}\right)+\ln\ln{x}-\ln\ln{y}>0;$$ 5. $x\geq e>y>1.$

In this case I am stuck.

Thank you!

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closed as off-topic by YCor, Carlo Beenakker, Gerald Edgar, Andrés E. Caicedo, Joseph Van Name Mar 8 at 21:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Carlo Beenakker, Andrés E. Caicedo, Joseph Van Name
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ A forum math.stackexchange.com is a right place for such type questions. $\endgroup$ – user64494 Mar 8 at 20:33
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    $\begingroup$ @user64494 It was there and it's unsolved there. $\endgroup$ – Michael Rozenberg Mar 8 at 20:35
  • $\begingroup$ Try setting x =y^t for t > 1; I think you will have an easier time of it when y is bigger than 1. This is the wrong forum for your question. Gerhard "Likes Simplifying With Term Rewriting" Paseman, 2019.03.08. $\endgroup$ – Gerhard Paseman Mar 8 at 20:39
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    $\begingroup$ This problem has a strong flavor of Olympiad, rather than research, mathematics. Yet, it seems very non-trivial to me, and there have been many other Olympiad-style problems posted and accepted at this site. I vote to re-open, and I'd be happy if somebody of those voted to close (or anybody else) would post a solution. $\endgroup$ – Seva Mar 9 at 7:38
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    $\begingroup$ @MichaelRozenberg Well done! $\endgroup$ – Todd Trimble 2 days ago
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Maple helps you in your case 5 through

DirectSearch:-GlobalOptima(x^(y^x)-y^(x^y), {x >= exp(1), y >= 1, y <= exp(1)});

[1.71828182845907, [x = 2.71828182845907, y = 1.], 174]

Addition. Mathematica confirms it by

NMinimize[{x^(y^x) - y^(x^y),  x >= Exp[1] && y >= 1 && y <= Exp[1]}, {x, y}]

{1.71828,{x->2.71828,y->1.}}

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    $\begingroup$ Why do you think that it helps? I don't think so. $\endgroup$ – Michael Rozenberg Mar 8 at 21:58
  • $\begingroup$ @Michael Rozenberg: The above results prove the inequality for your case 5: 1.71 is greater than zero. Don't hesitate to ask for further explanation in need. $\endgroup$ – user64494 Mar 9 at 4:44
  • $\begingroup$ These programs can get us a bug, which happens in very many cases. I think it's not a proof. $\endgroup$ – Michael Rozenberg Mar 9 at 4:49
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    $\begingroup$ @Michael Rozenberg: I prefer arguments over emotional words. $\endgroup$ – user64494 Mar 9 at 5:02

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