$\pi(x)\geqslant\frac{\log x}{2\log2}$ for all $x\geqslant2.$ Let $\pi$ be the prime counting function. Then

$\pi(x)\geqslant\log x/(2\log2)$ for all $x\geqslant2.$

Maybe I am missing something pretty evident, but, so far, I have proved that $\pi(x)\geqslant\log{\lfloor x\rfloor}/(2\log2)$ using a method Paul Erdős used to prove that there are infinitely many primes, however, I don't know how to prove the original inequality.
Any help is really appreciated!. 
 A: Take the first $j$ primes $2,3,\dots,p_{j}$ and define $N\left(x\right)=\left|\left\{ n\leq x:\, p\nmid n\,\forall p>p_{j}\right\} \right|$. If we write an $n$ in the form $$n=n_{1}^{2}m$$ with $m$ a squarefree number, we have $$m=2^{a_{1}}3^{a_{2}}\cdots p_{j}^{a_{j}}$$ where $a_{i}\in\left\{ 0,1\right\},\,i=1,\dots,j $. So we have $2^{j}$ possible choice of the exponents and so of differents $m$ and $n_{1}\leq\sqrt{n}\leq\sqrt{x}$, then we have no more than $\sqrt{x}$ choice of $n$. So $$N\left(x\right)\leq2^{j}\sqrt{x}.\tag{1}$$ Now take $j=\pi\left(x\right)$, then $N\left(x\right)=x$ and, by $(1)$, $$x\leq2^{\pi\left(x\right)}\sqrt{x}\Rightarrow\pi\left(x\right)\geq\frac{\log\left(x\right)}{2\log\left(2\right)}.$$ This also proves the bound $$\pi\left(2^{n}\right)\geq\frac{n}{2}$$ and I think it's the “simple” way that the commentator of Thomas Andwes question talking about. (You can find this proof in G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth Edition, Oxford Science Publications, 1996, p.16-17)
A: The basic idea how to show that, for any integer $n\ge 2$ we have
$$n\le 2^{\pi(n)}\sqrt n, \tag{1}$$
is to rewrite any number from $1,2,\dots,n$ in the form $j\cdot k^2$ where $j$ is square free. Then we have:

*

*At most $\sqrt n$ possibilities for $k$ since $k^2\le n$.

*At most $2^{\pi(n)}$ possibilities for $j$, since every square-free $j$ corresponds to a subset of $\{p_1,\dots,p_{\pi(n)}\}$.

From this we get $$2^{\pi(n)}\ge\sqrt n.\tag{2}$$
Still, for real $x$ between $n$ and $n+1$ we have $\pi(x)=\pi(n)$ and we only get
$$2^{\pi(x)}=2^{\pi(n)}\ge \sqrt n = \sqrt{\lfloor x \rfloor},$$
which is a bit weaker that the inequality that we want to prove.
Let us try to replace $(1)$ by a slightly finer estimate. We're still considering the numbers of the form $j\cdot k^2$ between one and $n$. But we notice that if $j$ is divisible by some prime $p>\frac n2$, then we actually must have $j=p$ and $k=1$. (Otherwise we get $j\cdot k^2>n$.) This leads to a new estimate
$$n\le 2^{\pi(n/2)}\sqrt n + \pi(n) - \pi(n/2). \tag{3}$$
This implies
$$\sqrt n\le 2^{\pi(n/2)} + \pi(n) - \pi(n/2). \tag{4}$$
Using the fact that for $2\le x\le y$ we have $2^y-2^x\ge 2(x-y)$ we get the following estimate for the RHS:
$$2^{\pi(n/2)} + \pi(n) - \pi(n/2) \le 
2^{\pi(n/2)} + \frac{2^{\pi(n)} - 2^{\pi(n/2)}}2 
\le 2^{\pi(n)-1}.$$
This leads to
$$\sqrt n \le 2^{\pi(n)-1}\\
2\sqrt n \le 2^{\pi(n)}$$
and finally
$$\sqrt{n+1} \le 2^{\pi(n)} \tag{5}$$
(Using $2\sqrt n\ge\sqrt{n+1}$ in the last step.)
The above argument works if $\pi(n/2)\ge2$, i.e., if $n\ge 6$. It remains to check that $(5)$ is true for $n=2,3,4,5$, too.
Once we have $(5)$, we see that for $n\le x<n+1$ we get
$$\sqrt x \le \sqrt{n+1} \le 2^{\pi(n)} \le 2^{\pi(x)}.$$
So we arrived at $$\sqrt x \le 2^{\pi(x)}\tag{6},$$ as desired.

This inequality was recently briefly discussed in chat. As you can see from the links there, this is mentioned in various places (several books, some other posts on this site). But usually only concentrating on the case of integers, without looking in detail whether we have the same inequality for real arguments.
Since the inequality is rather big overestimate, I am pretty sure that there are many various ways how to get $\sqrt{n+1} \le 2^{\pi(n)}$ instead of just $\sqrt{n} \le 2^{\pi(n)}$. I wanted to try to find at least one such argument.
