I would like to show that if
$$ c_1\frac{x}{\ln(x)} < \pi(x) < c_2 \frac{1}{\ln(x)} \hspace{1cm} \text{ (Chebyshev's Estimates)} $$
then
$$ \pi(2x) - \pi(x) > 1 \hspace{1cm} \text{ (Bertrand's Postulate)} $$
only if
$$ 2c_1 > c_2 $$
Although I suspect this question is about algebraic manipulation, the context is the Chebyshev estimates for the prime counting function being sufficiently narrow to prove Bertrand's Postulate (eg pdf) where:
$$\begin{align} c_1 &= \frac12 \ln(2) \\\\c_2 &= 6\ln(2)\end{align}$$
My own attempts only take me as far as the following:
The smallest difference $\pi(2x) - \pi(x)$ is when we consider the lower bound for $\pi(2x)$ and the upper bound for $\pi(x)$.
$$ c_1\frac{2x}{\ln(2x)} - c_2\frac{x}{\ln(x)} > 1 \hspace{1cm} \text{ (Bertrand's Postulate)}$$
From this I can't make progress.