Prove that if $e

Question

Prove that if $e<y<x$ then $x^{y}<y^{x}$

My try:
I tried to use Taylor's theorem: $$y^{x}-x^{y}=e^{xlny}-e^{ylnx}=1+xlny+o(xlny)-1-ylnx-o(ylnx)=$$ $$=lny^{x}-lnx^{y}+o(xlny)-o(ylnx)=ln\frac{y^{x}}{x^{y}}+o(xlny)-o(ylnx)$$Hovewer I have a problem to show that $y^{x}-x^{y}>0$ because I can't say that $ln\frac{y^{x}}{x^{y}}>0$ because then I use with theses.

Have you some idea?

Note that for $x, y > 0$, $x^y < y^x$ is equivalent to (by taking logs) $$y\ln x < x\ln y,$$ i.e. $$\frac{\ln x}{x} < \frac{\ln y}{y}.$$ So you end up being asked to show that $f(t):= \frac{\ln t}{t}$ is strictly decreasing on $(e,\infty)$. You could use calculus to help you show this. — Minus One-Twelfth, Apr 13 '19 at 12:11

score 4 · Accepted Answer · edited Mar 31 '24 at 15:41

4

$e^{x\ln(y)}<e^{y\ln(x)}$ is equivalent to ${{\ln(x)}\over x}<{{\ln(y)}\over y}$

Let $f(x)={{\ln(x)}\over x}$, $f'(x)={{1-\ln(x)}\over x^2}<0$ if $x>e$, so if $e<y<x$, $f(x)<f(y)$

edited Mar 31 '24 at 15:41

Robert Z

145,942

answered Apr 13 '19 at 12:14

Tsemo Aristide

87,475

score 2 · Answer 2 · answered Apr 13 '19 at 12:14

2

The hint:

Prove that $f(x)=\frac{x}{\ln{x}}$ increases for $x>e$.

answered Apr 13 '19 at 12:14

Michael Rozenberg

194,933

Prove that if $e

2 Answers2

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