TITLE: Proving $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$
QUESTION [14 upvotes]: How to prove this: $$\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$$
From Apostol's number theory text i know that $$\sum\limits_{p \leq x} \frac{1}{p} = \log{\log{x}} + A + \mathcal{O}\Bigl(\frac{1}{\log{x}}\Bigr)$$ But how can i use this to prove my claim.

REPLY [15 votes]: I wasn't sure that you wanted a proof that used that fact from Apostol.
One easy method not using the result you quote from Apostol is as follows:
$$
\sum_{p\le x} \frac{1}{\sqrt{p}} > \sum_{p \le x} \frac{1}{\sqrt{x}} = \frac{\pi(x)}{\sqrt{x}} > c \frac{\sqrt{x}}{\log{x}}
$$
where $c$ is a constant you can get from Chebyshev or Rosser and Schoenfeld, and maybe do a little computation to take care of small $x$, and you've got it.  This is a much better lower bound than what you are trying to prove.