, 2023

Note that for each $k = 1, 2, \dots, n$, we have
$$\frac{b_1}{a_1} + \frac{b_2}{a_2} + \dots + \frac{b_k}{a_k} \ge k \sqrt[k]{\frac{b_1}{a_1} \frac{b_2}{a_2} \dots \frac{b_k}{a_k}} \ge k,$$
which means that
$$s_k = \frac{b_1 - a_1}{a_1} + \frac{b_2 - a_2}{a_2} + \cdots + \frac{b_k - a_k}{a_k} = \frac{b_1}{a_1} + \frac{b_2}{a_2} + \cdots + \frac{b_k}{a_k} - k \geq 0.$$
For each $k = 1, 2, \dots, n$, we have
$$s_k - s_{k-1} = \frac{b_k - a_k}{a_k} \implies b_k - a_k = a_k s_k - a_k s_{k-1}$$
where $s_0 = 0$. Hence,
$$\begin{aligned} & (b_1 - a_1) + (b_2 - a_2) + \cdots + (b_n - a_n) \\ &= a_1 s_1 - a_2 s_1 + a_2 s_2 - a_3 s_2 + \cdots + a_n s_n - a_n s_{n-1} \\ &= s_1(a_1 - a_2) + s_2(a_2 - a_3) + \cdots + s_{n-1}(a_{n-1} - a_n) + s_n a_n \geq 0, \end{aligned}$$
as desired.