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?