Prime counting function; is it true that $\pi(n - m) \geq \pi(n) -\pi(m)$?

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 $n \geq 2.$

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 $n \le 300$.

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.