$a \leq b$, is $a^b \leq b^a$ correct? I have just thought about an interview question, it was maybe asked previously, but I thought about it myself.

Consider, $a \leq b$, is $a^b \leq b^a$ correct? Justify.

I thought about solving it in the following manner, but don't end with a conclusive result.
Let's consider:
$$\begin{align}
a^b &\leq b^a \\
e^{\ln(a^b)} &\leq e^{\ln(b^a)} \\
\ln(a^b) &\leq \ln(b^a) \\
b\ln(a) &\leq a\ln(b)
\end{align}
$$
I know that $\ln(a) \leq \ln(b)$, but cannot conclude from there.
Do you have any suggestion?
 A: Here is a slightly more systematic way of looking at it. 
You want to check whether 
$$a \le b \implies \frac{\log a}{a} \le \frac{\log b}{b}$$
This would be true if the function $f: \mathbb{R} \to \mathbb{R}$ 
mapping $x \mapsto \frac{\log x}{x}$ were an increasing function. 
Now, if you know a little calculus, we can look at the derivative to 
get a bit of a clue. It's hopefully easy for you to check that 
$$f'(x) = \frac{1-\log x}{x^{2}}.$$
Now the relationship between increasing/decreasing functions and their 
derivatives is the sign of the derivative. It is also hopefully easy 
to see that 
$$f'(x) \text{ is } \begin{cases} 
\text{positive } & \text{ if } x < e \\
\text{zero } & \text{ if } x = e \\
\text{negative } & \text{ if } x > e
\end{cases}$$
So, $f$ is increasing, stationary, and decreasing on the respective intervals. 
Oh no! If $f$ is decreasing, then our desired property will fail. So pick some 
$e < a < b$, and you should get a counterexample. Indeed, 
$$3 \le 4 \text{ but } 3^{4} > 4^{3}$$
as (un)desired. 
A: Counter-example:
$2\le 5$, yet $\;2^5>5^2$.
A: When looking for counterexamples, I recommend trying the simplest cases first.
Are the signs of $a, b$ restricted in any way?  If not, try $a = -1, b = 2$.
