Showing that $x^{\frac{1}{x}}-1-\frac{\ln(x)}{x}<\frac{1}{x\ln(x)}$ for all $x>1$ When trying to come up with an asymptotic formula for the partial sums $\sum_{n\leq x}(\sqrt[n]{n}-1)$, I came across the following inequality which seems to be to holding for all $x>1$ :
\begin{equation}\tag{$*$}
x^{\frac{1}{x}}-1-\frac{\ln(x)}{x}<\frac{1}{x\ln(x)}.
\label{eq:special}
\end{equation}
I am unable to prove this, but numerical evidences suggest that it is correct.
I have only been able to prove a weaker inequality for all $x>1$
$$x^{\frac{1}{x}}-1-\frac{\ln(x)}{x}<\frac{1}{ex}.$$
This is obtained by rewriting $x^{\frac{1}{x}}-1-\frac{\ln(x)}{x}=\frac{1}{x}\int_{1}^{x}\frac{t^{1/x}-1}{t}dt$ and noticing that for all $x>1$ and $t>0$, we have (by computing the derivative of $f(t)=\frac{t^{1/x}-1}{t}$ and solving $f'(t)=0$),
$$\frac{t^{1/x}-1}{t}\leq\frac{(x-1)^{x-1}}{x^x}=\frac{1}{x-1}\left(1-\frac{1}{x}\right)^{x}\leq\frac{1}{(x-1)e}.$$
Something that also comes to mind when dealing with \eqref{eq:special} is to write
\begin{equation}
x^{\frac{1}{x}}-1-\frac{\ln(x)}{x}=\exp{\left(\frac{\ln(x)}{x}\right)}-1-\frac{\ln(x)}{x}=\sum_{k=2}^{+\infty}\frac{1}{k!}\frac{\ln^{k}(x)}{x^{k}}
\end{equation}
and try to bound the infinite sum by $\frac{1}{x\ln(x)}$. This however didn't help me prove the inequality.
Does anyone have any hints on how to prove \eqref{eq:special}.
 A: Fact 1: Let $C = \mathrm{e}^{2 + 1/\mathrm{e}} - \mathrm{e}^2 - \mathrm{e} \approx 0.567393943$. It holds that, for $x > 1$,
$$x^{\frac{1}{x}}-1-\frac{\ln(x)}{x} \le C\frac{\ln^2 x}{x^2}.$$
It is easy to prove that $C\frac{\ln^2 x}{x^2} < \frac{1}{x\ln(x)}$ for $x > 1$ (omitted). We are done.
$\phantom{2}$
Proof of Fact 1: Let $y = \frac{\ln x}{x}$. It is easy to prove that $0 < y \le \frac{1}{\mathrm{e}}$ for $x > 1$ (omitted).
We need to prove that, for $0 < y \le \frac{1}{\mathrm{e}}$,
$$\frac{\mathrm{e}^y - 1 - y}{y^2} \le C.$$
Let $f(y) = \frac{\mathrm{e}^y - 1 - y}{y^2}$. We have $f'(y) = \frac{2-y}{y^3}(\frac{2+y}{2-y} - \mathrm{e}^y)$.
It is easy to prove that $f'(y) > 0$ for $0 < y \le \frac{1}{\mathrm{e}}$ (omitted).
Thus, $f(y) \le f(\frac{1}{\mathrm{e}}) = \mathrm{e}^{2 + 1/\mathrm{e}} - \mathrm{e}^2 - \mathrm{e}$. We are done.
A: Because the graph of the function $x \mapsto 1/x$ lies above the
tangent at $(1, 1),$
\begin{gather*}
u < \int_{1 - u/2}^{1 + u/2}\frac{dx}x = \ln\frac{1 + u/2}{1 - u/2},
\text{ therefore } e^u < \frac{2 + u}{2 - u} \quad (0 < u < 2), \\
\text{therefore } e^u - 1 - u < \frac{u^2}{2 - u} \quad (0 < u < 2),
\\ \text{therefore } x^{1/x} - 1 - \frac{\ln x}x
< \frac{(\ln x)^2}{x(2x - \ln x)} \quad (x > 1).
\end{gather*}
The required inequality follows if we can prove
$(\ln x)^3 + \ln x < 2x$ for all $x > 1.$ Equivalently,
$t^3 + t < 2e^t$ for all $t > 0.$ But the derivative of $t^3e^{-t}$
is $t^2(3 - t)e^{-t},$ and the derivative of $te^{-t}$ is
$(1 - t)e^{-t},$ therefore $t^3e^{-t} \leqslant (3/e)^3$ and
$te^{-t} \leqslant 1/e$ for all $t > 0.$ Therefore, because
$e > 2.7,$
$$
(t^3 + t)e^{-t} \leqslant \left(\frac3e\right)^3 + \frac1e
< \frac{1000}{729} + \frac{10}{27} = \frac{1270}{729} < 2
$$
for all $t > 0,$ as required.
A: As suggested in the comments, I will answer my own question.
We have for all $x>1$, as $\vert\frac{\ln(x)}{x}\vert<1$,
\begin{equation}
x^{1/x}-1-\frac{\ln(x)}{x}=\sum_{k=2}^{+\infty}\frac{1}{k!}\left(\frac{\ln(x)}{x}\right)^{k}\leq \frac{1}{2}\sum_{k=2}^{+\infty}\left(\frac{\ln(x)}{x}\right)^{k}=\frac{1}{2}\frac{\ln^{2}(x)}{x^2-x\ln(x)}.
\end{equation}
It remains to show that $\frac{1}{2}\frac{\ln^{2}(x)}{x^2-x\ln(x)}\leq \frac{1}{x\ln(x)}$ for every $x>1$, or equivalently to show that $\frac{1}{2}\ln^{3}(x)+\ln(x)\leq x$ for every $x>1$.
To do so, let $f(x)=\frac{1}{2}\ln^{3}(x)+\ln(x)$ and $g(x)=x$. Observe that $f(1)=0$, $g(1)=1$, and that $f'(x)\leq g'(x)$ for all $x>1$.
Indeed, $f'(x)=\frac{3}{2}\frac{\ln^{2}(x)}{x}+\frac{1}{x}$ and some relatively simple calculations show that for all $x>1$,
\begin{equation}
f'(x)=\frac{3}{2}\frac{\ln^{2}(x)}{x}+\frac{1}{x}\leq \frac{3+\sqrt{3}}{\exp(1+\frac{1}{\sqrt{3}})}< 1=g'(x).
\end{equation}
In fact, the simpler estimate
\begin{equation}
x^{1/x}-1-\frac{\ln(x)}{x}\leq\frac{1}{2}\frac{\ln^{2}(x)}{x^2-x\ln(x)}=\frac{\ln^2(x)}{x^2}\frac{1}{2}\left(\frac{1}{1-\frac{\ln(x)}{x}}\right)\leq \frac{\ln^2(x)}{x^2}\frac{1}{2}\left(\frac{1}{1-\frac{1}{e}}\right)
\end{equation}
is much better for large $x$.
