= Solution
Put $N=\lfloor x\rfloor$. Comparing the <harmonic number> with an <integral> gives $H_N=\sum_{n=1}^N1/n>\int_1^{N+1}dt/t=\log(N+1)>\log x$. By the <Fundamental theorem of arithmetic>, the finite <Euler product> $\prod_{p\leq x}(1-p^{-1})^{-1}$ expands as the sum of $1/n$ over positive integers whose <prime factors> are at most $x$. It therefore includes every term of $H_N$, and
$$
\log H_N\leq-\sum_{p\leq x}\log(1-p^{-1})
<\sum_{p\leq x}\frac1p+\frac12\sum_{p\leq x}\frac1{p(p-1)}
<\sum_{p\leq x}\frac1p+\frac12.
$$
The first strict inequality is the <geometric-series bound for the logarithmic remainder>; the last uses the <telescoping series> $\sum_{n=2}^\infty1/[n(n-1)]=1$. Thus the <prime reciprocal lower bound> is
$$
\boxed{\sum_{p\leq x}\frac1p>\log H_N-\frac12>\log\log x-\frac12.}
$$
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