Upper bound of natural logarithm

Question

I was playing looking for a good upper bound of natural logarithm and I found that

$$\ln x \le x^{1/e}$$

apparently works: Can someone give me a formal proof of this inequality?

Answer 1

Consider the fuction $$f(x)=\log( x) -x^{1/e}$$ Its derivative $$f'(x)=\frac{e-x^{\frac{1}{e}}}{e x}$$ cancels for $x=e^e$ and, for this value $f(x)=0$; the second derivative test shows that this is a maximum($f''(e^e)=-e^{-1-2 e}$). Then $$\log( x) \leq x^{1/e}$$ is always satisfied.