= Solution
Write the <Chebyshev theta function> as
$$
\vartheta(x)=\sum_{\substack{p\leq x\\p\text{ prime}}}\log p.
$$
Taking <natural logarithms> in part a gives
$$
2n\log2-\log(2n+1)
\leq\sqrt{2n}\log(2n)+\vartheta(2n),
$$
and hence
$$
\vartheta(2n)\geq
2n\log2-\sqrt{2n}\log(2n)-\log(2n+1).
$$
Because $\sqrt n\log n/n\to0$, the right-hand side is at least $n/2$ for every sufficiently large <integer> $n$.
Now let $x$ be sufficiently large and set $n=\lfloor x/2\rfloor$ using the <floor function>. Then $2n\leq x$ and $n\geq x/3$, so the monotonicity of $\vartheta$ gives
$$
\boxed{
\sum_{\substack{p\leq x\\p\text{ prime}}}\log p
=\vartheta(x)\geq\vartheta(2n)
\geq\frac n2\geq\frac{x}{6}\geq\frac{x}{12}.}
$$
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