I'm not sure if it's what you're looking for, but the proof of Bertrand's postulate is somewhat related to the proof of Chebyshev's theorem, which says that $\pi(x) = \Theta(x/\log x)$; both proofs rely on careful estimation of the central binomial coefficients $\binom{2n}{n}$.
To prove Chebyshev's theorem, one can start by finding the estimate
$$
\frac{2^{2n}}{2n+1} < \binom{2n}{n} < 2^{2n},
\tag{1}
$$
then using it to conclude that $\vartheta(x)/x < 4\log 2$ and hence
$$
\limsup_{x \to \infty} \frac{\pi(x)}{x/\log x} = \limsup_{x \to \infty} \frac{\vartheta(x)}{x} \leq 4 \log 2,
$$
and $\psi(x)/x > (x-2)\log 2/x - \log(x+1)/x$ and hence
$$
\liminf_{x \to \infty} \frac{\pi(x)}{x/\log x} = \liminf_{x \to \infty} \frac{\psi(x)}{x} \geq \log 2,
$$
where $\vartheta$ and $\psi$ are Chebyshev's functions.
To prove Bertrand's postulate, one can improve the estimate in $(1)$ to
$$
\frac{2^{2n}}{2\sqrt{n}} < \binom{2n}{n} < \frac{2^{2n}}{\sqrt{2 n}},
\tag{2}
$$
which can be used to show that
$$
\vartheta(2n) - \vartheta(n) > \left(\frac{2n}{3}-1\right)\log 2-\left(\frac{\sqrt{2n}+1}{2}\right)\log 2 > 0$$
for $n \geq 2^6$. This is exactly Bertrand's postulate for $n \geq 2^6$. The remaining cases are checked by hand.
My reference for these proofs is Chandrasekharan's Introduction to Analytic Number Theory.
Best Answer
The result is fairly elementary. Lets prove it now:
Recall some common definitions: Let $\theta(x)=\sum_{p\leq x} \log p$, let $\Lambda(n)$ be the Von Mangoldt lambda function and let $\psi(x)=\sum_{n\leq x} \Lambda(n)$
Then as $\log n =\sum_{d|n} \Lambda(n)$ we see that $$\sum_{n\leq x} \log n=\sum_{n\leq x}\sum_{d|n} \Lambda(d)=\sum_{d\leq x} \Lambda(d)\lfloor x/d\rfloor =x\sum_{d\leq x} \frac{\Lambda(d)}{d}+O\left(\psi(x) \right)$$
Lemma 1: $\theta(x)\leq 3x$
The proof of this is postponed until the end.
Now, since $\theta(x)=\psi (x)+O(\sqrt{x})$, we have $\psi(x)=O(x)$. Combining this with the fact that $\sum_{n\leq x} \log n= x\log x + O(x)$ we see that $$\sum_{d\leq x} \frac{\Lambda(d)}{d}=\log x+O(1).$$ Since the sum over all the prime powers of $\frac{\Lambda(d)}{d}$ is bounded by a constant, we conclude $$\sum_{p\leq x}\frac{\log p}{p}=\log x+O(1).$$
Hope that helps,
Proof of Lemma 1:
Consider the binomial coefficient $$\binom{2N}{N}.$$ Notice that every prime in the interval $(N,2N]$ appears in the numerator. Then $$\prod_{N<p\leq2N}p\leq\binom{2N}{N}$$ Since $$\binom{2N}{N}\leq(1+1)^{2N}=4^{N},$$ we see that $$\prod_{N<p\leq2N}p\leq4^{N},$$ and hence $\theta(2N)-\theta(N)\leq N\log4.$ Consider $N$ of the form $N=2^{r}.$ Then we get the list of inequalities: $$\theta\left(2\right)-\theta\left(1\right)\leq\log4$$ $$\theta\left(4\right)-\theta\left(2\right)\leq2\log4$$
$$\theta\left(8\right)-\theta\left(4\right)\leq4\log4$$
$$\theta(2^{r})-\theta\left(2^{r-1}\right)\leq2^{r-1}\log4$$ $$ \theta\left(2^{r+1}\right)-\theta\left(2^{r}\right)\leq2^{r}\log4.$$
Summing all of these yields $$\theta\left(2^{r+1}\right)-\theta\left(1\right)\leq\left(2^{r}+\cdots+1\right)\log4\leq2^{r+1}\log4$$ for each $r$. For every $x$ in the interval $(2^{r},2^{r+1}]$ we have $\theta(x)\leq\theta(2^{r+1})$ so that $$\theta(x)\leq2^{r+1}\log4\leq x\cdot2\log4<3x$$ since $x>2^{r}$. Thus the result is proven.