Minimum Argument Difference to Make the Lower Bound > the Upper Bound Assume $g$ is a function that grows asymptotically as 
$$
g(n) \in\frac n {log(n)} + O(\sqrt n),\,n \in \Bbb N\tag1
$$
I wish to find $h(n)$ such that
$$
g(n) \le g(n+h(n)).
$$
i.e. Given the bounds
$$
\begin{align}
u(n)=\frac n {log(n)} + \sqrt{n} \tag 2 \\
l(n)=\frac n {log(n)} - \sqrt{n} \tag 3
\end{align}
$$
what $h(n)$ would give
$$
u(n)\le l(n+h(n)) \tag 4
$$
Since $h(n)$ would simply be the difference of the inverse functions of $(2)$ and $(3)$, my approach so far has been to look for these. The inverse of $f(n)=n/log(n)$ is given by
$$
f^{-1}(n)=e^{-W(-1/n)} \tag 5
$$
which has the multivalued Lambert W function, where the principal branch for reals stops at $e$. The analytic continuation is
$$
f^{-1}(n) = n \log(n \log(n \log(...\,\,,\,\, n>e \tag 6
$$ 
but I haven't found a way to express the inverse of the bounding functions. Maybe I need a taylor series or some integral to approximate these?  

Proof of $(6)$
The inverse of $y = {x / log(x)}$ is
$$
x = {y / log(y)}.\tag7
$$
Assume $(7)$ can be equated to
$$
y=x\,log(x\,log(x\,log(x\,...\tag8
$$
The expression inside the first log function is $y$, so we have
$$
y=x\,log(y).\tag9
$$
Dividing through by $log (y)$ confirms the original assumption (for $x>e$). $(8)$ returns the upper branch value (the area of interest), whereas $(5)$ returns only the principal value.  

Addendum
Apparently, $(1)$ is weaker than I intended. How could it be made stronger, such that $(2)$ and $(3)$ are strict bounds? My guess is that substituting $O()$ with something like
$$
o(x^{\frac 1 2 - \epsilon})\tag{10}
$$
would eliminate the implied constant of $O()$, and reflect my intention that
$$
\forall n \in \Bbb N, l(n) < g(n) < u(n)\tag{11}
$$
and $(4)$ holds such that
$$
\lim_{n\to \infty} \frac {l(n+h(n))}{u(n)} \to 1^+.\tag{12}
$$

It turns out that the optimal choice of $h(n)$ depends on the constant implicit in the $O(\sqrt{n})$ term, so redefine
$$
u(n) = \frac{n}{\log n} + c\sqrt{n}
$$
and
$$
l(n) = \frac{n}{\log n} - c\sqrt{n},
$$
where $c > 0$ is fixed.  Assuming $h = o(n)$, solving the equation
$$
u(n) = l(n+h)
$$
asymptotically for $h$ yields
$$
h(n) \sim 2c\sqrt{n}\log n. \tag{1}
$$
Further, it's possible to show that
$$
\lim_{n\to\infty} u(n) - l(n+2c\sqrt{n}\log n) = \infty \tag{2}
$$
and that, for any $\epsilon > 0$,
$$
\lim_{n\to\infty} u(n) - l(n+(2+\epsilon)c\sqrt{n}\log n) = -\infty. \tag{3}
$$
Consequently, if
$$
l(n) \leq g(n) \leq u(n)
$$
for $n$ large enough then for any $\epsilon > 0$ there exists an $n_0(\epsilon)$ such that
$$
g(n) \leq g(n+(2+\epsilon)c\sqrt{n}\log n) \tag{4}
$$
for $n \geq n_0(\epsilon)$.
(Note that we cannot take $\epsilon = 0$ in general, since any function $g(n)$ which oscillates between $l(n)$ and $u(n)$ will satisfy
$$
g(n) > g(n+2c\sqrt{n}\log n)
$$
infinitely often.  For a concrete example, take $g(n) = n/\log n + c\sqrt{n}\sin n$.)

Proof of $(1)$.
If we expand
$$
\begin{align}
l(n+h) &= \frac{n+h}{\log(n+h)} - c\sqrt{n+h} \\
&= \frac{n+h}{\log n + \log(1 + h/n)} - c\sqrt{n+h} \\
&= \frac{n+h}{\log n} \frac{1}{1 + \frac{\log(1+h/n)}{\log n}} - c\sqrt{n}\sqrt{1+h/n} \\
&= \frac{n+h}{\log n} \left[1 + O\left(\frac{h}{n\log n}\right)\right] - c\sqrt{n} \left[1 + O\left(\frac{h}{n}\right)\right] \\
&= \frac{n+h}{\log n} + O\left(\frac{h}{(\log n)^2}\right) - c\sqrt{n} + O\left(\frac{h}{\sqrt{n}}\right),
\end{align}
$$
then the equation $l(n+h) - u(n) = 0$ becomes
$$
\frac{h}{\log n} + O\left(\frac{h}{(\log n)^2}\right) - 2c\sqrt{n} + O\left(\frac{h}{\sqrt{n}}\right) = 0. \tag{5}
$$
A dominant balance is achieved if
$$
\frac{h}{\log n} \asymp 2c\sqrt{n},
$$
so it follows that $h \sim 2c\sqrt{n} \log n$.

Proof of $(2)$ and $(3)$.
We know from $(5)$ that, for $h = o(n)$,
$$
u(n) - l(n+h) = 2c\sqrt{n} - \frac{h}{\log n} + O\left(\frac{h}{(\log n)^2}\right) + O\left(\frac{h}{\sqrt{n}}\right).
$$
By taking $h = (2+\varepsilon)c\sqrt{n}\log n$ with $\epsilon > 0$ we see that
$$
u(n) - l(n+h) \sim -\epsilon c \sqrt{n} \to -\infty,
$$
which proves $(3)$.
Proving $(2)$ is only slightly more difficult; we need to expand the first term of $l(n+h)$ to one more order.  I'll omit the calculations here and just state the result that, for $h = o(n)$,
$$
u(n) - l(n+h) = 2c\sqrt{n} - \frac{h}{\log n} + \frac{h}{(\log n)^2} +  O\left(\frac{h^2}{n(\log n)^2}\right) + O\left(\frac{h}{\sqrt{n}}\right).
$$
By taking $h = 2c\sqrt{n}\log n$ we see that
$$
u(n) - l(n+h) \sim \frac{2c\sqrt{n}}{\log n} \to \infty,
$$
which proves $(2)$.
