Prime counting function; is it true that $\pi(n - m) \geq \pi(n) -\pi(m)$? Let $\pi$ be the prime counting function, and $m, n$ positive integers, $n > m > 1$.
Is it proven that $\pi(n - m) \geq \pi(n) -\pi(m)$ ?
 A: Note that the edited form is equivalent to the question "Does every sequence of $m>1$ consecutive positive integers contain at most $\pi(m)$ primes?"
This is still false when $n = m+1$ is a prime, but perhaps you can patch this further by assuming $m < n-1.$
Answer to original version:
Obviously not.  Just try some values.
In particular, just by using $m=1$ inductively, this would show
$\pi(n) < 0$ for all $n \geq 2.$
A: If we allow $m = 1$, then this statement is $\pi(n-1) > \pi(n)$, which is always false.
If we require $m > 1$, these are still equal infinitely often. For any prime $p$, choose $m = p-1$ and $n = p+1$. Then clearly $\pi(n) - \pi(m) = 1 = \pi(2) = \pi(n-m)$.
However, if we require $n > m > 1$ and relax the inequality to $\pi(n-m) \ge \pi(n)-\pi(m)$, this might hold for all $n,m$. Mathematica informs me it does for all $n \le 300$.
A: The modified version (with the additional requirement that $m<n-1$) is just the second Hardy–Littlewood conjecture, which has not been proven. In fact, it has been proven that the second Hardy–Littlewood conjecture is inconsistent with the prime tuples conjecture.
