I want to show that $$\sum\limits_{p\leq X,\textrm{ }p \textrm{ prime}}\frac{1}{p}>\log\log X-\frac{1}{2},$$ for all $X\geq1$.
I have already been able to show that $$\prod\limits_{p\leq X,\textrm{ }p\textrm{ prime}}(1-p^{-s})^{-1}\geq\sum\limits_{n\leq X}n^{-s}$$ for all $s\in\mathbb{R}_{>0}$, and that $$\sum\limits_{n\geq X}\frac{1}{n}>\log X.$$ (These two identities are previous parts of the exercise.)
Now I tried going about the problem as follows: $$\sum\limits_{p\leq X,\textrm{ }p \textrm{ prime}}\frac{1}{p}=\sum\limits_{n\leq X}\frac{1}{n}-\sum\limits_{n\leq X,\textrm{ }n\textrm{ not prime}}\frac{1}{n}.$$ Next I conjecture that $$\sum\limits_{n\leq X,\textrm{ }n\textrm{ not prime}}\frac{1}{n}>\sum\limits_{n\leq X}\frac{1}{2n},$$ by drawing a graph. However, I'm not 100% sure that's correct, and neither how to proceed from here.
$$\sum_{p\leq X}\frac{1}{p}>\sum_{p\leq X}-\frac{4}{5p^2}-\log\left(1-\frac{1}{p}\right)=-\sum_{p\leq X}\frac{4}{5p^2}+\log\prod_{p\leq X}\left(1-\frac{1}{p}\right)^{-1}$$ is greater than
$$\log\sum_{n\leq X}\frac{1}{n}-\frac{4}{5}\sum_{p}\frac{1}{p^2}>\log\log X+\frac{4}{5}\sum_{p}\log\left(1-\frac{1}{p^2}\right)=\log\log X-\frac{4}{5}\log(\zeta(2)) $$ so $$ \sum_{p\leq X}\frac{1}{p}>\log\log X-\frac{2}{5}.$$