Weak versions of Bertrand's postulate We are interested in the following statement: 
For each $n>1$ and $x>2$ there is at least one prime $p$ satisfying $x<p<n x$.
For $n=2$ we get precisely the Bertrand's postulate which is true. As corollaries, the statements for arbitrary $n\geqslant 2$ are true. However, I am interested maybe there exist independent proofs (prossibly short and elementary) for some of the cases $n>2$.
Thanks for your help.  


Theorem [Chebyshev]: There are $A,B\in\mathbb R$ such that:
    $$A\cdot \frac x{\log x}<\pi(x)<B\cdot \frac x{\log x}$$

Given such $A$ and $B$, choose $n\in\mathbb N$ big enough such that $\frac n{1+2\log n}\geq\frac BA$. Then, for $x\geq 2$:
$$
\pi(nx) 
> A\cdot\frac{nx}{\log (nx)}
\geq B\cdot\frac{(1+2\log n)\cdot nx}{n\log (nx)}
\geq B\cdot\frac{(1+\frac{\log n}{\log x})\cdot nx}{n\log (nx)}
$$
$$
= B\cdot\frac{\frac{\log x + \log n}{\log x}\cdot x}{\log (nx)}
= B\cdot\frac{\frac{\log (nx)}{\log x}\cdot x}{\log (nx)}
= B\cdot\frac{x}{\log x}
> \pi(x)
$$
So there is a prime between $x$ and $nx$.

Of course, there are different proofs to Chebyshev's theorem, depending on how tight you want the bounds to be. For instance, there is an elementary proof yielding $A=\frac{\log 2}{4}$ and $B=8\log 2$. Then $\frac BA=32$ and $n=419$ does the trick. 
