Singapur

Note that for any positive numbers $a, b$ between $p$ and $q$, we may assume (due to symmetry) that $0 < p \le a \le b \le q$. We have
$$\frac{a}{b} + \frac{b}{a} \le \frac{p}{q} + \frac{q}{p} \iff \frac{a^2 + b^2}{ab} \le \frac{p^2 + q^2}{pq} \\ \iff p^2 ab + q^2 ab - a^2 pq - b^2 pq = (bq - ap)(aq - bp) \ge 0$$
Since the last inequality is true, the first is true as well. Thus
$$\text{LHS} = \left(\frac{a_1}{a_9} + \frac{a_9}{a_1}\right) + \dots + \left(\frac{a_4}{a_6} + \frac{a_6}{a_4}\right) + \frac{a_5}{a_5} \le 1 + \frac{4(p^2 + q^2)}{pq}$$