$x^e \le e^x$, is this true? 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.
 A: Take logarithm, you can since $x>0$. Now your inequality becomes:
$$e \log x \le x \log(e)$$
Or
$$\frac{\log x}x \le \frac 1e$$
To prove this, you just have to study the function
$$f(x)=\frac{\log x}x$$
Differentiate:
$$f'(x)=\frac{1-\log x}{x^2}$$
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
$$f(e)=\frac{1}{e}$$
Hence your inequality is true.
A: 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.
A: 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 $x=\mathrm{e}$.
