For all positive real numbers $a_1, a_2, \dots, a_n$, prove that $$ \frac{a_1\! +\! a_2}{2} \cdot \frac{a_2\! +\! a_3}{2} \cdot \dots \cdot \frac{a_n\! +\! a_1}{2} \leq \frac{a_1\!+\!a_2\!+\!a_3}{2 \sqrt{2}} \cdot \frac{a_2\!+\!a_3\!+\!a_4}{2 \sqrt{2}} \cdot \dots \cdot \frac{a_n\!+\!a_1\!+\!a_2}{2 \sqrt{2}}.$$