Show that $x^e\le e^x$ for all $x\gt 0$ and $x \in \mathbb {R}$

Question

I understand that this question may seem quite simple, but although I can see different ways of showing this, I don't understand how it follows from the context I was given (i.e why the second part of the question begins with "hence"). It may also help to bear in mind that I have only just started teaching myself calculus and so far I have only covered differential calculus up to the level taught in secondary schools.

The question is in two parts (I understand that to answer (i) you simply find $f'(x)$ and evaluate $f'(e)$ to show that it is equal to zero and is therefore, since there is only one stationary point which is a maximum point as implied by the question, the maximum point):

$f(x) = {\ln x\over x}$, $x\gt 0$

(i) Show that the maximum point on the graph of $y = f(x)$ occurs at the point $\left(e,\frac{1}{e}\right)$.

(ii) Hence, show that $x^e\le e^x$ for all $x\gt 0$

Any help would be greatly appreciated.

Answer 1

Consider the function $F(x)= x^{\frac 1x}$. We claim that $F(x)$ has a maximum at $x=e$. Indeed your $f(x)$ is $\log F(x)$ so this follows from what you have proved. But this then means that $$x^{\frac 1x}\leq e^{\frac 1e}\implies x^e\leq e^x$$

Answer 2

By i) you have determined

$f(x) = \frac {\ln x}x \le \frac 1e; \forall x > 0$.

So $e \ln x \le x$

So $e^{e\ln x} \le e^x$

And $e^{e\ln x}=(e^{\ln x})^e = x^e \le e^x$.

==

Part i) is a matter of taking $f'(x)$ and setting to zero.

$f'(x) = (\frac {\ln x}x)' = \ln x(-\frac 2{x^2}) + \frac 1x*\frac 1x = \frac{1- \ln x}{x^2}$

$f'(x) = 0 \implies$

$\ln x = 1$

$x = e$

$f''(x) = -\frac 1{x^2} - 2\ln x \frac 1{x^3} < 0$ for $x = e$ so $x=e; f(x) = \frac 1e$ is maximum and for all $x > 0; \frac {\ln x}x \le \frac 1e$.