10. Let positive numbers $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ satisfy: $a_{1}, a_{2}, a_{3}$ form an arithmetic sequence with a common difference of $b_{1}$, $b_{1}, b_{2}, b_{3}$ form a geometric sequence with a common ratio of $a_{1}$, and $a_{3}=b_{3}$. Find the minimum value of $a_{3}$, and determine the value of $a_{2} b_{2}$ when $a_{3}$ takes its minimum value.
Given $a_{3}=a_{1}+2 b_{1}, b_{3}=b_{1} a_{1}^{2}$, let $a_{3}=b_{3}=\lambda$, then from $\lambda=a_{1}+2 b_{1} \Rightarrow b_{1}=\frac{\lambda-a_{1}}{2}$, substituting into $b_{3}=b_{1} a_{1}^{2} \Rightarrow \lambda=\frac{\left(\lambda-a_{1}\right) a_{1}^{2}}{2}=\frac{\left(2 \lambda-2 a_{1}\right) a_{1}^{2}}{4} \leqslant \frac{1}{4}\left(\frac{2 \lambda}{3}\right)^{3}=\frac{2 \lambda^{3}}{27}$ $\Rightarrow \lambda^{2} \geqslant \frac{27}{2} \Rightarrow \lambda \geqslant \frac{3 \sqrt{6}}{2}$, equality holds when $2 \lambda-2 a_{1}=a_{1} \Rightarrow a_{1}=\frac{2}{3} \lambda=\sqrt{6}$. At this point, $b_{1}=\frac{\sqrt{6}}{4}, a_{2}=\frac{5 \sqrt{6}}{4}, b_{2}=\frac{3}{2} \Rightarrow a_{2} b_{2}=\frac{15 \sqrt{6}}{8}$.