A one-line lower bound for prime counts

 (Zhao Yu here). Today someone spoke about this short proof of a lower bound of the number of primes, which I found it too nice to not share.

Theorem:

Let $\pi(n)$ be the number of primes $\le n$, then $$\pi(n)\gtrsim \frac{n}{\log_2 n}.$$

Proof:

Let $P(x)=x^n(1-x)^n=a_{2n}x^{2n}+a_{1}x^{2n-1}+\cdots+a_{0}$. Then,

$$\frac{1}{2^{2n}}\ge\int_0^1P(x)dx = \sum_{i=0}^{2n} \frac{a_i}{i+1}=\frac{c}{lcm(1,\cdots,2n+1)}\ge \frac{1}{(2n+1)^{\pi(2n+1)}}.$$

The first inequality uses AM-GM on $x^n(1-x)^n$, and in the last inequality we note that $c$ has to be a positive integer.