Prove the inequality $e^x \geq x^e$ for $x > 0$

Question

Prove that $e^x \ge x^e$ for $x \gt 0$

I applied the natural logarithm to simplify the function and I get $$\frac{x}{\ln x}\ge e$$

How to solve these types of problems?

Answer 1

You are alsmot there.

Study the function $f(x)=\dfrac{\ln x}{x}$

Then $f'(x)=\dfrac{1}{x^2}(1-\ln x)$

$f'(x)>0$ for $x<e$ and $f'(x)<0$ for $x>e$

But $f(e)=\dfrac 1e$...

Answer 2

$x \geq e\ln x \iff x - e\ln x \geq 0$. $f(x) = x - e\ln x \to f'(x) = 1-\dfrac{e}{x}=\dfrac{x-e}{x}\geq 0 \iff x \geq e$. Thus if $x \geq e, f(x) \geq f(e) = 0 \to x \geq e\ln x \to e^x \geq x^e$, and if $x \leq 0, f(x) \geq f(e) = 0 \to e^x \geq x^e$.

Answer 3

$e^{x/e}$ is convex and its tangent at $x=e$ is $y = x$, hence $e^{x/e} \ge x$

Answer 4

You could define the function \begin{equation}f(x)=e^x-x^e \end{equation} and show that $f'(x)$ is zero at $x=1,e$ and go forth in terms of the minima of $f(x)$. With this, however, you will also have to show that $f(x)$ doesn't achieve a minimum at $\pm\infty$...