Prove that, for all $x \geqslant 2$, we have

$$\sum_{p \leqslant x} \frac{1}{p}>\log \log x-\frac{1}{2}$$

[You may assume that, for $0<u<1$,

$$-\log (1-u)-u<\frac{u^{2}}{2(1-u)} .$$