= Solution
Use <sprinkling of a binomial random graph> to write $G(n,p)=G_1\cup G_2$, where the rounds are independent,
$$
p_1=10\frac{\log n}{n},
\qquad
p_2=c\frac{\log n}{n},
$$
and $C$ is chosen large enough that $1-p=(1-p_1)(1-p_2)$. By the given theorem, $G_1$ has a <Hamilton cycle> $v_1v_2\cdots v_nv_1$ <with high probability>.
Condition on such a cycle. For each $3\leq\ell\leq n-1$, every chord $v_iv_{i+\ell-1}$ closes one of the two paths around the Hamilton cycle into a cycle of length $\ell$. There are at least $n/2$ distinct candidate chords, so the probability that $G_2$ supplies none is at most
$$
(1-p_2)^{n/2}\leq e^{-p_2n/2}=n^{-c/2}.
$$
A <union bound> over the fewer than $n$ lengths shows that all these cycles occur simultaneously <with high probability> when $c>4$. The Hamilton cycle itself supplies length $n$, so $G$ is <pancyclic graph>[pancyclic].
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