$\pi(x)\geqslant\frac{\log x}{2\log2}$ 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!.

Solution 1:

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)