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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?

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    What do you mean: *Is there any other way*? That is, without exponentiating or without breaking into cases? At any rate...just *divide*!2012-12-26

3 Answers 3

1

It is well known that $1+x\leq e^x$, thus:

$x $\log(x)<\log(e^x)=x$

4

If $f(x)=\ln x-x\ln x$ then $f(1)=f'(1)=0$ and $f''(x)\le0$ for all $x>0$.

3

The functions $\ln x$ and $x\ln x$ are continuous. So $f(x)=x\ln x -\ln x$ can only change sign where $f(x)=0$. This happens at $x=1$. Now use "test" values, say $x=1/e$ and $x=e$. (Possibly very close to your cases analysis.)