Show $\ln x \le x \ln x$, when $x > 0$.
Using exponentiation, this is
$$
x \le x^x.
$$
Doing a case by case analysis for the cases
\begin{align}
(1) &: x \in (0,1), \\
(2) &: x = 1, \\
(3) &: x > 1,
\end{align}
gives the results.
Is there another way?