TITLE: prove that $a^b\ge{b}^a$ where $a\le{b}$.
QUESTION [5 upvotes]: prove that $a^b\ge{b}^a$ for all $a,b\ge3$. given that $a\le{b}$.
I was trying to solve the question by graph. Can anyone help me please?

REPLY [2 votes]: It is enough to show that the function $f\left( t \right) = \frac{{\ln t}}{{t +
\ln t}}$, is decreasing for all $t\ge 3$. Since $$
f'\left( t \right) = \frac{{\left( {t + \ln t} \right)\frac{1}{t} - \left( {1 + \frac{1}{t}} \right)\ln t}}{{\left( {t + \ln t} \right)^2 }} \le 0
$$
for all $t\ge e$. Therefore, for any $x,y\ge 3$ such that if $y\ge x$ then $f(x)\ge f(y)$, i.e.,
\begin{align}
&\frac{{\ln x}}{{x + \ln x}} \ge \frac{{\ln y}}{{y + \ln y}}
\\
&\Rightarrow y\ln x + \ln x\ln y \ge x\ln y + \ln x\ln y
\\
&\Rightarrow y\ln x \ge x\ln y
\\
&\Rightarrow x^y  \ge y^x. 
\end{align}
You may also, consider the decreasing function $f\left( t
\right) = \frac{{\ln t}}{t}$, for all $x\ge e$.