Solving this inequality: $e^x \geq x^e$

Question

Show that $e^x \geq x^e$ for $ 0 \lt x \lt \infty $.

I tried to apply the normal logarithm here, which yields $x \geq e\times \ln(x)$ Still, I am kind of stuck here, anyone mind giving me a hand?

Answer 1

Hint

Consider the function $f(x) = e^x - x^e$ and find its minimum over positive $x$. If you do it correctly (e.g. solve $f'(x) = 0$, etc) you will find that it is non-negative. Hence, the claim will follow.

UPDATE

Easier to take logs: $e^x \ge x^e$ iff $x \ge e \ln x$, and now look at $f(x) = x - e \ln x$, which should be elementary

Answer 2

Hint: $e^{x/e}$ is convex and $y=x$ is its tangent at $x=e$.

Answer 3

Consider the function $f(x)=x-eln(x)$. $f'$ exists, $$ f'(x) = 1-\frac{e}{x}. $$ That function has a root, Which are easily found: $x=e$. Now, $f'$ is negative in $e-h$ and positive in $e+h$ with appropriate $h$, so that critical point is a minimum. When plugged into the original function, we have $0$, so the argument is indeed true.