For any integers $a < b$, we have $ab \le (a+1)(b-1)$. Hence to maximise the product $a_1a_2a_3 \cdot \ldots \cdot a_m$, we make the terms as close to each other as possible. If $m$ is even, this is achieved by setting half to $9$ and the other half to $11$, so that we have $(a_1a_2a_3 \cdot \ldots \cdot a_m)^{\frac{1}{m}} = (9^{\frac{m}{2}} 11^{\frac{m}{2}})^{\frac{1}{m}} = 3\sqrt{11}$. If $m$ is odd, we have two cases to check: Case 1: There are $\frac{m+1}{2}$ of $9$, $\frac{m-3}{2}$ of $11$, and $1$ of $12$. In this case $(a_1a_2a_3 \cdot \ldots \cdot a_m)^{\frac{1}{m}} = \left(9^{\frac{m+1}{2}} \cdot 11^{\frac{m-3}{2}} \cdot 12\right)^{\frac{1}{m}} = 3\sqrt{11} \cdot \left(\frac{1296}{1331}\right)^{\frac{1}{2m}} < 3\sqrt{11}$. Case 2: There are $1$ of $8$, $\frac{m-3}{2}$ of $9$, and $\frac{m+1}{2}$ of $11$. In this case $(a_1a_2a_3 \cdot \ldots \cdot a_m)^{\frac{1}{m}} = \left(8 \cdot 9^{\frac{m-3}{2}} \cdot 11^{\frac{m=1}{2}}\right)^{\frac{1}{m}} = 3\sqrt{11} \cdot \left(\frac{704}{729}\right)^{\frac{1}{2m}} < 3\sqrt{11}$. So we are done. $\blacksquare$