Was Landau the first to prove that there is a prime on $\bigl(x,\frac65x\bigr)$?

In his *Handbuch* $\!^1$ discussing the limit

$$\lim_{n\to\infty} \bigl(\pi\bigl((1+\epsilon)x\bigr)-\pi(x)\bigr)=\infty $$

he seems to say that in the next chapter he will prove the relation for all $\epsilon > \frac{1}{5}.$ On the face of it this is the answer and Jitsuro Nagura's proof$^2$ for $\epsilon \geq 1/5$ is not only an improvement but uses methods that Landau felt were exhausted by his proof, which is what I take from
Landau's "*da man doch nicht auf diesem elementaren Wege das Ziel erreichen kann...*"

In the following section (21) of the following chapter (50) he derives a constant from sums involving the $\psi$ function and in section 22 proves Bertrand's Postulate. In 23 entitled "*Weitere Verengerung der Schranken*" [further narrowing of bounds] he performs another series of manipulations of $\psi(x)$ and derives that

$$(A)\hspace{7mm} \limsup_{x\to\infty}\frac{\psi(x)}{x}\leq\frac{171\cdot 6}{175\cdot 5}a \approx 1.08028 $$ in which $a\approx 0.92129\dots$ and on the same page finds

$\psi(x)\geq ax+o(x) \approx 0.92129 x + o(x).$

So for comparison,

Nagura obtains

$0.916x-2.318 < \psi(x) < 1.086x$

and Landau obtains

$0.92129x + o(x) < \psi(x) < 1.08028 x.$

Landau halts his proof after (A), concluding that (A) "*besser als $\frac{6a}{5}$ ist,*" leaving the reader I think a bit of work.

I haven't done the calculations yet but if as a quick check we substitute $0.93 x\leq \psi(x)\leq 1.085x$ into Nagura's expressions$^3$ for the difference $\vartheta\bigl(\frac{n+1}{n}x\bigr)-\vartheta(x)$ we get positive values for $ n = 5$ and $x$ near $5000$.

So my question is whether Landau could have said $\epsilon\geq \frac{1}{5}$ using his own methods?

Landau and Nagura worked in different circumstances so this is really just curiosity about the nature of Landau's result.

$^1$ *Handbuch* (1909)p. 87.

$^2$ Nagura, On the Interval Containing At Least One Prime Number (1952).

$^3$ These values are a little worse than those Landau obtained assuming Landau's expression for the difference may involve some loss of numerical leverage over the expression in Nagura's paper.