= Solution
\b[The statement is true.] For $n\geq2$, being in every pairwise intersection is equivalent to being in all the events:
$$
\bigcap_{i<j}(A_i\cap A_j)=\bigcap_{k=1}^n A_k.
$$
Indeed every index occurs in at least one pair. Apply the <union bound> to the complements of these $\binom n2$ pairwise intersections to obtain
$$
P\left(\bigcap_{k=1}^n A_k\right)
\geq1-\sum_{i<j}\{1-P(A_i\cap A_j)\}
=\sum_{i<j}P(A_i\cap A_j)-\binom n2+1>0.
$$
This is <positive common intersection from pair-intersection probabilities>. It remains valid for $n=2$, where it reduces to the asserted positivity of the single pairwise intersection.
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