TITLE: Conjecture: $\pi(x)\ge \pi\circ\pi(x)+\pi\circ\pi\circ\pi(x)+\cdots$
QUESTION [16 upvotes]: $x\ge 13\implies\pi(x)\ge \pi\circ\pi(x)+\pi\circ\pi\circ\pi(x)+\cdots$

Can this be proved?

REPLY [7 votes]: The inequality is trivially true because from the prime number theorem, the LHS is asymptotic to $\frac{x}{\log x}$ where as the RHS is asymptotic to $\frac{x}{\log^2 x}$ so clearly this is a very weak inequality and is has to be true for all $x$ greater than some minimum value which in this case turns to be 13. 
What would make the inequality non-trivial is if the LHS and RHS are of the same asymptotic order.  I present a stronger form of the above inequality.
Let $s(x) = \pi(\pi(x)) + \pi(\pi(\pi(x))) + \ldots $. 
Pierre Dusart has proved that
$$
\pi(x) \ge \frac{x}{\log x - 1}, \ x > 5393
$$
$$
\pi(x) \le \frac{x}{\log x - 1.1}, \ x > 60,184
$$
From these two inequalities and  little algebraic manipulation we obtain
$$
\frac{x}{(\log x - 1)(\log x - 2)} < s(x) < \frac{x}{(\log x - \log\log x)(\log x - \log\log x - 1)}
$$
or in the slightly weaker but simpler form 
$$
\frac{x}{(\log x - 1)^2} < s(x) < \frac{x}{(\log x - \log\log x - 1)^2}
$$
for all $x > 60,184$. Clearly the RHS is less than $\pi(x)$ so the weaker original inequality follows.