# Prime counting function of $x \lt c \frac{x}{\log x}$ proof clarification

I am reading Kenneth Ireland and Michael Rosen's A Classical Introduction to Modern Number Theory, and I'm having trouble with a step in one of their proofs.

The proof in question is of Corollary 1 in chapter 2, section 4. The corollary is stated as follows:

There is a positive constant $$c_{1}$$ such that $$\pi(x) < c_{1}x/$$log $$x$$ for $$x \ge 2$$.

It goes without saying, but for the benefit of readers new to number theory, $$\pi(x)$$ is the number of primes less than or equal to $$x$$ and is known as the prime counting function. The proof employs a function on the natural numbers defined as $$\theta(x) = \sum_{p \le x}$$ log $$p$$, the sum being over all primes at most x. The proof begins as follows:

$$\theta(x) \ge \sum^{p \le x}_{p > \sqrt{x}}$$ log $$p \ge$$ (log$$\sqrt{x})(\pi(x) - \pi(\sqrt{x})) \ge$$ (log$$\sqrt{x})\pi(x) - \sqrt{x}$$log$$\sqrt{x}$$.

The first inequality follows from the definition of $$\theta(x)$$, the next follows from the fact that log $$p >$$ (log$$\sqrt{x})$$ and $$\pi(x) - \pi(\sqrt{x})$$ is the number of primes $$p$$ that satisfy $$\sqrt{x} < p \le x$$. However, the last inequality seems to imply that $$\pi(\sqrt{x}) \ge \sqrt{x}$$ if you multiply out the (log$$\sqrt{x}$$) term, and that can't be true.

I'd like to know if there's some logic I'm missing here, like maybe they aren't multiplying out the (log$$\sqrt{x}$$) term and are instead using some trick to get that last inequality. Perhaps they're using one of the earlier propositions about $$\pi(x)$$, namely Proposition 2.4.1: $$\pi(x) \ge$$ log(log$$x$$) or Proposition 2.4.2: $$\pi(x) \ge$$ log$$x/(2$$log$$2)$$ - though I'm having trouble spotting how they might be using them.

Edit: Silly mistake, going to leave this question up as a reminder for myself to stay humble.

Your misunderstanding is simple. They actually use $$\pi(\sqrt{x})\le \sqrt{x}$$, which is obviously true and equivalent to $$-\pi(\sqrt{x})\ge -\sqrt{x}$$. Since the term has a negative sign, this is the correct result.