Problem 1919

Firstly, we prove that
$$\max_{1 \le k \le n} a_k < 2,\ \text{ and } \max_{1 \le k \le n} b_k < \frac{2}{n-1}.$$
Let $L = \max_{1 \le k \le n} a_k$. From (b) and $\sum_{k=1}^{n} b_k = 1$, we get $L^2 \le 1 + L$, so $L < 2$.
Let $b_m = \max_{1 \le k \le n} b_k$. Then by using $2b_k \ge b_{k-1} + b_{k+1}$, it is easy to see that
$$b_k \ge \begin{cases} \frac{(k-1)b_m + (m-k)b_1}{m-1}, & 1 \le k \le m, \\ \frac{(k-m)b_n + (n-k)b_m}{n-m}, & m \le k \le n. \end{cases}$$
Since $b_1 > 0$ and $b_m > 0$, so
$$b_k > \begin{cases} \frac{k-1}{m-1}b_m, & 1 \le k \le m, \\ \frac{n-k}{n-m}b_m, & m \le k \le n. \end{cases}$$
It follows that
$$\begin{aligned} 1 &= \sum_{k=1}^{n} b_k = \sum_{k=1}^{m} b_k + \sum_{k=m+1}^{n} b_k \\ &> \frac{1}{m-1} \left( \sum_{k=1}^{m} (k-1) \right) b_m + \frac{1}{n-m} \left( \sum_{k=m+1}^{n} (n-k) \right) b_m \end{aligned}$$
$$= \frac{m}{2}b_m + \frac{n-m-1}{2}b_m = \frac{n-1}{2}b_m.$$
So $b_m < \frac{2}{n-1}$, that is $\max_{1 \le k \le n} b_k < \frac{2}{n-1}$.

Now let $f_0 = 1$, $f_k = 1 + \sum_{i=1}^k a_i b_i$, $k = 1, 2, \dots, n$. Then $f_k - f_{k-1} = a_k b_k$, and from (b) we have $a_k^2 \le f_k$, i.e. $a_k \le \sqrt{f_k}$, $k = 1, 2, \dots, n$.
Since $\max_{1 \le k \le n} a_k < 2$, so
$$f_k - f_{k-1} = a_k b_k \le b_k \sqrt{f_k}$$
and
$$f_k - f_{k-1} < 2b_k.$$
Thus, for $1 \le k \le n$,
$$\begin{align*} \sqrt{f_k} - \sqrt{f_{k-1}} &< b_k \cdot \frac{\sqrt{f_k}}{\sqrt{f_k} + \sqrt{f_{k-1}}} \\ &= b_k \left( \frac{1}{2} + \frac{f_k - f_{k-1}}{2(\sqrt{f_k} + \sqrt{f_{k-1}})^2} \right) \\ &< b_k \left( \frac{1}{2} + \frac{2b_k}{2(\sqrt{f_k} + \sqrt{f_{k-1}})^2} \right) \\ &< b_k \left( \frac{1}{2} + \frac{b_k}{4} \right) \\ &< \left( \frac{1}{2} + \frac{1}{2(n-1)} \right) b_k. \end{align*}$$
Hence, summing from $k=1$ to $n$,
$$\begin{align*} a_n \le \sqrt{f_n} < \sqrt{f_0} + \sum_{k=1}^{n} \left( \frac{1}{2} + \frac{1}{2(n-1)} \right) b_k \\ = \frac{3}{2} + \frac{1}{2(n-1)}. \end{align*}$$
Let $n \to +\infty$, we obtain $a_n \le \frac{3}{2}$.

When $a_k = 1 + \frac{k}{2n}$, $b_k = \frac{1}{n}$, $k = 1, 2, \dots, n$, we have
$$a_k^2 = \left(1 + \frac{k}{2n}\right)^2 \le 1 + \sum_{i=1}^k \frac{1}{n}\left(1 + \frac{i}{2n}\right)$$
Hence the maximum value is $\frac{3}{2}$.