$ \pi(n) > \frac{n}{22 \ln(n)} $ simple proof? What is the easiest way to show that
$$ n > 2 $$
$$ \pi(n) > \frac{n}{2 \ln(n)} $$
Where $\pi(n)$ is the prime counting function. 
I read a proof of the PNT with the zeta function but this statement is much weaker !!
What is the shortest proof ? 
The simplest ?
The most elementary ?
Do we use results from Mertens ? ( $\Pi ( 1 - 1/p) $ or $ \sum 1/p $ )
Do we need to use results of Mertens ?
Do we need to estimate $\sum \ln(p) $ ?
How about the even weaker
$$ \pi(n) > \frac{n}{22 \ln(n)} $$
Is that even easier ? Or not ? 
 A: Too long for a comment:
In Section 4.5 of Apostol's Introduction to Analytic Number Theory (page 82-84), we have

Theorem 4.6 For every integer $n\ge2$, we have
$${1\over6}{n\over\log n}\lt\pi(n)\lt6{n\over\log n}$$

The proof is elementary. This takes care of the OP's question with $22$ in the denominator, but leaves open the question whether there's a simple proof if you replace Apostol's $6$ with a $2$. Apostol does, however, introduce the theorem saying, "Although better inequalities can be obtained with greater effort (see [60]) the following theorem is of interest because of the elementary nature of its proof." Reference [60] is:

Rosser, J. Barkley, and Schoenfeld, Lowell (1962) Approximate formulas
  for some functions of prime number theory. Illinois J. Math.,
  6:69-94; MR 25, #1139.

A: I think, the proof given by Dusard in his thesis is the best one:
we have
$$
\pi(x)  \ge \frac{x}{\log(x)}\left( 1+\frac{1}{\log(x)}+\frac{1.8}{\log^2(x)}\right)
$$
for all real $x\ge 32299$, and this bound is sharp. For all $x\ge x_0$  we obtain the bounds in general, and then we test for $\frac{1}{2}$ or $\frac{1}{22}$ for all $x\le x_0$ by a direct computation.
There is also an elementary proof by Chebychev, see here, with the variant Daniel is suggesting above. But it is also a bit tedious to do this and a general results perhaps seems to be useful then.
