I know that $\sum_{p \leq N} \frac{1}{p} \geq \log\log N -1 $
However, want to show that $\sum_{p \leq x} \frac{1}{p} \geq \log\log x -1 $.
If let $N=[x]$, then we get a bound for x, i.e. $N \leq x <N+1$. However, from that all I can seem to get is this $\sum_{p \leq N+1} \frac{1}{p} \geq \log\log(N+1) -1 >\log\log x-1 $
Can't seem to be able to conclude that. As we have $\frac{1}{N+1}$
Well, in your other thread we got $\displaystyle \sum_{p \leq N} \frac{1}{p} \geq \log\log(N+1) -1\ \ $ (with $N \to N+1$)
and this is enough to get $\displaystyle \sum_{p \leq x} \frac{1}{p} \geq \log\log x -1\ \ $ (for $N=\lfloor x \rfloor$)