Firstly note that it is impossible fpr $p$ to take the value $0$, since it is a divisor, then we have that $0 < p\leq a_i\leq q$. Then we have that $a_i\leq q$ and $\frac{1}{a_i} \leq \frac{1}{p}$ by condition, thus the LHS of the inequality is less than or equal to $\frac{9p}{q}$. Then the result is true if and only if $$9p^2\leq pq+4p^2+4q^2 \iff 0\leq (q-p)(4q+5p)$$which is evidently true, as desired.

WLOG $a_9\geq a_8 \geq a_7 \geq ... \geq a_1$ Then by Re-Arrangement inequality we have $$ LHS \leq \frac{a_1}{a_1}+\frac{a_2}{a_2}+...+\frac{a_9}{a_9}=9$$by AM GM $$ p^2+q^2 \geq 2 qp \rightarrow RHS \geq 8+1=9$$ Equality happens when $a_1=a_2=...=a_9=p=q$

Construct the decreasing sequence $b_1\ge b_2\ge\dots\ge b_9$ whose elements are those in $\{a_1,a_2,\dots,a_9\}$. By rearrangement inequality we have that \[ \sum_{i=1}^9 a_{i}\cdot \frac{1}{a_{n-i}} \le \sum_{i=1}^9 b_{i}\cdot \frac{1}{b_{n-i}} \le \sum_{i=1}^9 \frac{q}{p}=9\cdot \frac{q}{p}. \]Note that we have $9\cdot \frac{q}{p}\le 1+\frac{4p^2+4q^2}{pq}\iff 9p^2\le pq+4p^2+4q^2$, which is of course true, since $p\le q$. So we can state that $LHS\le 9\cdot \frac{q}{p}\le 1+\frac{4p^2+4q^2}{pq}$, which is what we wanted to show.