Question 135, Given a positive integer $m(m \geq 3)$, 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.
Question 135, Solution: Let $a_{1}=a, b_{1}=b$, then $a, b \in \mathbb{R}^{+}$, and according to the conditions: $a_{m}=a+(m-1)b, b_{m}=b \cdot a^{m-1}$. Therefore,
$$
\begin{array}{l}
a+(m-1) b=b \cdot a^{m-1} \\
\Rightarrow b=\frac{a}{a^{m-1}-(m-1)}
\end{array}
$$

Let $a^{m-1}-(m-1)=t$, then $a=(t+m-1)^{\frac{1}{m-1}}$, hence
$$
a_{m}=b \cdot a^{m-1}=\frac{a}{a^{m-1}-(m-1)} \cdot a^{m-1}=\frac{(t+m-1)^{m-1}}{t}=\left(\frac{(t+m-1)^{m}}{t^{m-1}}\right)^{\frac{1}{m-1}}
$$

According to the AM-GM inequality:
$$
\begin{array}{l}
(m-1)^{2} t^{m-1} \leq\left(\frac{(m-1)^{2}+(m-1) t}{m}\right)^{m}=\left(\frac{m-1}{m}\right)^{m} \cdot(t+m-1)^{m} \\
\Rightarrow \frac{(t+m-1)^{m}}{t^{m-1}} \geq \frac{(m-1)^{2}}{\left(\frac{m-1}{m}\right)^{m}}=\frac{m^{m}}{(m-1)^{m-2}}
\end{array}
$$

Substituting (2) into (1) we know: the minimum value of $a_{m}$ is $\left(\frac{m^{m}}{(m-1)^{m-2}}\right)^{\frac{1}{m-1}}$, which occurs when $t=(m-1)^{2}$. At this point, $\frac{a_{1}}{b_{1}}=\frac{a}{b}=a^{m-1}-(m-1)=(t+m-1)-(m-1)=t=(m-1)^{2}$.

Note: Let $f(x)=\frac{(x+m-1)^{m}}{x^{m-1}}$, then $f(x)=\frac{m \cdot(x+m-1)^{m-1} \cdot x^{m-1}-(x+m-1)^{m} \cdot(m-1) \cdot x^{m-2}}{x^{2(m-1)}}$, hence $f(x)=0 \Leftrightarrow$ $x=(t-1)^{2}$, indicating that $f(x)$ is monotonically decreasing on $\left(0,(t-1)^{2}\right)$ and monotonically increasing on $\left((t-1)^{2},+\infty\right)$. Therefore, $f_{\min }(x)=f\left((t-1)^{2}\right)=\frac{m^{m}}{(m-1)^{m-2}}$, hence the minimum value of $a_{m}$ is $\left(\frac{(t+m-1)^{m}}{t^{m-1}}\right)^{\frac{1}{m-1}}$.