 Solution

Use two history D flip-flops. Write $a=a_n$, $b=a_{n-1}$ and $c=a_{n-2}$ for the signals just before the active clock edge; update the history by $b^{+}=a$ and $c^{+}=b$. Two or more of the three samples are one exactly when some pair are both one, so the majority function isThree two-input AND gates and one three-input OR gate implement this disjunctive normal form. If a registered output is wanted, feed the expression to the D input of another D flip-flop and capture it at the same edge as the history update. All right-hand sides use the pre-edge values, so no extra sample of delay is introduced.

The three terms are necessary in a minimal sum of products: the input triples $110$, $101$ and $011$ must each be covered, and a single product covering two of them would also cover a triple having only one one. Thus the minimum is three products, each with two Boolean literals.