# Prove that $x^{y^x}>y^{x^y}$ for $x>y>0.$ [closed]

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 $$y^x or $$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!

## closed as off-topic by YCor, Carlo Beenakker, Gerald Edgar, Andrés E. Caicedo, Joseph Van NameMar 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.

• A forum math.stackexchange.com is a right place for such type questions. – user64494 Mar 8 at 20:33
• @user64494 It was there and it's unsolved there. – Michael Rozenberg Mar 8 at 20:35
• 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. – Gerhard Paseman Mar 8 at 20:39
• 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. – Seva Mar 9 at 7:38
• @MichaelRozenberg Well done! – Todd Trimble 2 days ago

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]

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