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Is $x^e \le e^x$ for any given $x > 0$, where $f\colon (0, \infty) \to R$ and $f(x) = \ln(x) / x$

I don't know exactly how should I demonstrate this.

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What has $f$ to do with the question? However, the title inequality has already been treated here. – egreg Dec 26 '13 at 21:23
what does $f(x)$ have to do with the equality you're trying to show? – hunter Dec 26 '13 at 21:23
$e^t\geq t+1:$ let $t=\frac{x}{e}-1$ and raise to the $e^{\text{th}}$ power. – Julien Godawatta Dec 26 '13 at 21:31
up vote 8 down vote accepted

Take logarithm, you can since $x>0$. Now your inequality becomes:

$$e \log x \le x \log(e)$$


$$\frac{\log x}x \le \frac 1e$$

To prove this, you just have to study the function

$$f(x)=\frac{\log x}x$$


$$f'(x)=\frac{1-\log x}{x^2}$$

Now, if $x<e$, then $f'(x)>0$, and $f$ is increasing. And for $x>e$, $f'(x)<0$ and $f$ is decreasing. Thus $f$ has a global maximum at $x=e$, and


Hence your inequality is true.

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The claim is that $$ e \log(x) \leq x $$

or equivalently

$$ \log(x) \leq \frac{x}{e} $$

When $x = e$ we get equality, the line $y = \frac{x}{e}$ is the tangent line here, and the graph of the logarithm is concave down, so this is true.

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That f(x) was in the statement, I think it was given so I can demonstrate that inequality using the function derivative. – Adrian George Dec 26 '13 at 21:28

Yes it is.

It is equivalent to showing that $f(x)=x^{1/x}\le \mathrm{e}^{1/\mathrm{e}}$, for all $x>0$.

Note that $f'(x)=x^{1/x}\frac{1-\ln x}{x^2}$, which vanishes only at $x=\mathrm{e}$, and it is positive in $(0,\mathrm{e})$, while it is negative in $(\mathrm{e},\infty)$. This show that the maximum of $f$ is achieved at $x=\mathrm{e}$.

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