10. Given a positive integer $m(m \geqslant 3)$, let the positive term arithmetic sequence $\left\{a_{n}\right\}$ and the positive term geometric sequence $\left\{b_{n}\right\}$ satisfy: the first term of $\left\{a_{n}\right\}$ equals the common ratio of $\left\{b_{n}\right\}$, the first term of $\left\{b_{n}\right\}$ equals the common difference of $\left\{a_{n}\right\}$, and $a_{m}=b_{m}$. Find the minimum value of $a_{m}$, and determine the ratio of $a_{1}$ to $b_{1}$ when $a_{m}$ takes the minimum value.
Let $\{a_{n}\}$ have a common difference of $d$, and $\{b_{n}\}$ have a common ratio of $q$, then $a_{n}=q+(n-1) d, b_{n}=d q^{n-1}$.
According to the problem, let $a_{m}=b_{m}=\lambda$, then $\lambda=q+(m-1) d \Rightarrow d=\frac{\lambda-q}{m-1}$, substitute into
$$
\begin{array}{l}
\lambda=d q^{m-1}=\frac{(\lambda-q) q^{m-1}}{m-1}=\frac{(m-1)(\lambda-q) q^{m-1}}{(m-1)^{2}} \\
\leqslant \frac{1}{(m-1)^{2}}\left[\frac{(m-1)(\lambda-q)+(m-1) q}{m}\right]^{m}=\frac{(m-1)^{m-2} \lambda^{m}}{m^{m}} \\
\Rightarrow \lambda^{m-1} \geqslant \frac{m^{m}}{(m-1)^{m-2}} \Rightarrow \lambda \geqslant \sqrt[m-1]{\frac{m^{m}}{(m-1)^{m-2}}}, \text { equality holds when }(m-1)(\lambda-q)=q \\
\Rightarrow q=\frac{(m-1) \lambda}{m}, \text { at this time } d=\frac{\lambda-q}{m-1}=\frac{\lambda}{m(m-1)}, \text { so } \frac{a_{1}}{b_{1}}=\frac{q}{d}=(m-1)^{2} .
\end{array}
$$