prove that $a^b\ge{b}^a$ where $a\le{b}$. 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?
 A: Consider the function
$$f(x)=\frac{x}{\ln(x)}$$ 
We have
$$f '(x)=\frac{\ln(x)-1}{\ln(x)^2}$$
which if positive for $x\ge 3$
So, f(x) is strictly increasing for $x \ge 3$
So, due to $a\le b$, we get
$$\frac{a}{\ln(a)}\le \frac{b}{\ln(b)}$$
So, we have
$$a \ln(b) \le b \ln(a)$$
because we can multiply with $\ln(a) \cdot \ln(b)$, since this is positive.
Taking exp on both sides leads to
$$b^a\le a^b$$
A: 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 $x\ge e$. 
A: Rearrange the inequality (separate the variables!) to make it claim monotonicity of a certain function, which can be proven using elementary calculus (compute derivative, determine sign).
