Bertrand's postulate proof Regarding http://michaelnielsen.org/polymath1/index.php?title=Bertrand%27s_postulate
I think the last inequality should be $4^{n/3}\le(2n+1)(2n)^{\sqrt{2n}}$. But even when the RHS is decreased from $(2n+1)(2n)^{\sqrt{2}n}$, the RHS still dominates the LHS for $n>>0$ (you can check with wolfram alpha:
http://www.wolframalpha.com/input/?i=%282n%2B1%29%282n%29^{sqrt%282n%29}+-+4^{n%2F3} ). 
This doesn't give any contradiction.
 A: Wolframalpha only shows values for small $n$, a smooth graph around $0$ doesn't necessarily predict what happens at $\infty$. Note that the standard plots in WA are for $ -15 \leq n \leq 15$ which is very far from $n > > 0$.
About that inequality:
$$4^{n/3}\le(2n+1)(2n)^{\sqrt{2n}} \Leftrightarrow n \frac{\ln(4)}{3} \leq \ln(2n+1)+\sqrt{2n} \left(\ln(2n)\right) (*)$$ 
Now use that $\ln(n) < < n^{\frac{1}{4}}$ for $n$ large enough to get a  contradiction. 

Added To clarify a little on the choice $n> 2048$ in the proof. Note that while WA shows that the graph seems to fail around $460$, a finite WA plot is NOT a proof, since the graph could also come back up. And WA was not available to Chebashev.
One needs to effectively prove that $(*)$ fails for $n > N$. Here is a proof for $N=2048$, the person which wrote the wiki article probably had something similar in mind.
I am going to switch from $\ln$ to $\log_2$. 
Claim: For $n >2048$ we have
$$ n \frac{\log_2(4)}{3} >  \log_2(2n+1)+\sqrt{2n}\left(\log_2(2n)\right) (*)$$ 
Proof:
Let $k$ be the integer so that $k \leq \log_2(n) < k+1$.
Then
$$ \frac{2}{3} n \geq \frac{2^{k+1}}{3}$$
$$\log_2(2n+1)+\sqrt{2n}\left(\log_2(2n)\right) <  k+2 + 2^{\frac{k+2}{2}}\cdot (k+2)$$
let $m=k+2$. If we can show that for all $m  \geq 11$ we have
$$\frac{2^m}{6} \geq m(2^{m/2}+1) $$
we are done. 
This is equivalent to 
$$2^m \geq 6m (2^{m/2}+1)$$
As $6m(2^{m/2}+1)< 6m( (2^{m/2}+2^{m/2})=12m2^{m/2}$$
this follows immediately from 
$$2^{m/2} \geq 12 m $$
which is an easy induction problem.
