Logarithmic inequality: $-\ln(x) ≤ (x)^{-\frac{1}{e}}$

Question

I'm trying to prove the following inequality: $$- \ln (x) \leq (x)^{-\frac{1}{e}} $$

over $[0, 1]$. I'm not sure how to move forward. I know that the equality is at $x = e^{-e}$. Any help is greatly appreciated.

Answer 1

If we set $x=e^{-t}$, we just have to show that $$ \forall t\geq 0,\qquad t\leq \exp\frac{t}{e} \tag{1}$$ but that is trivial by convexity: $f(t)=\exp\frac{t}{e}$ is a convex function on $\mathbb{R}$, and the equation of the tangent line at $x=e$ is exactly $g(t)=t$.

Answer 2

Let $f(x)=-\ln x-x^{-\frac1e}$. Then $$ f'(x)=\frac1{ex}(-e+x^{-\frac1e})$$ and hence $f(x)$ is decreasing if $x>e^{-e}$ and increasing if $0<x<e^{-e}$. So if $x>e^{-e}$, then $f(x)<f(e^\frac1e)=0$. Thus the solution of $\ln x<x^{-\frac1e}$ is $x>e^{-e}$.