Taiwan IMO Selection Camp, 2022

$$s_k = a_1 + a_2 + \dots + a_k \quad \text{and} \quad b_k = \frac{a_k s_{k-1}^2}{1 - a_k},$$
with the convention that $s_0 = 0$. Note that $b_k$ is exactly a summand in the sum we need to estimate. We shall prove the inequality
$$b_k < \frac{s_k^3 - s_{k-1}^3}{3}. \qquad (1)$$
Indeed, it suffices to check that
$$\begin{align*} (1) &\Longleftrightarrow 0 < (1-a_k)((s_{k-1}+a_k)^3 - s_{k-1}^3) - 3a_k s_{k-1}^2 \\ &\Longleftrightarrow 0 < (1-a_k)(3s_{k-1}^2 + 3s_{k-1}a_k + a_k^2) - 3s_{k-1}^2 \\ &\Longleftrightarrow 0 < -3a_k s_{k-1}^2 + 3(1-a_k)s_{k-1}a_k + (1-a_k)a_k^2 \\ &\Longleftrightarrow 0 < 3(1-a_k - s_{k-1})s_{k-1}a_k + (1-a_k)a_k^2 \end{align*}$$

$$b_1 + b_2 + \dots + b_n < \frac{s_n^3 - s_1^3}{3} = \frac{1}{3},$$
as desired.

First, let us define
$$S(a_1, \dots, a_n) := \sum_{k=1}^{n} \frac{a_k}{1-a_k} (a_1 + a_2 + \dots + a_{k-1})^2.$$
For some index $i$, denote $a_1 + \dots + a_{i-1}$ by $s$. If we replace $a_i$ with two numbers $a_i/2$ and $a_i/2$, i.e. replace the tuple $(a_1, \dots, a_n)$ with $(a_1, \dots, a_{i-1}, a_i/2, a_i/2, a_{i+1}, \dots, a_n)$, the sum will increase by
$$\begin{align*} S(a_1, \dots, a_{i-1}, a_i/2, a_i/2, a_{i+1}, \dots, a_n) - S(a_1, \dots, a_n) &= \frac{a_i/2}{1-a_i/2} \left(s^2 + \left(s + a_i/2\right)^2\right) - \frac{a_i}{1-a_i} s^2 \\ &= a_i \frac{(1-a_i)(2s^2 + sa_i + a_i^2/4) - (2-a_i)s^2}{(2-a_i)(1-a_i)} \\ &= a_i \frac{(1-a_i-s)sa_i + (1-a_i)a_i^2/4}{(2-a_i)(1-a_i)}, \end{align*}$$
which is strictly positive. So every such replacement strictly increases the sum. By repeating this process and making maximal number in the tuple tend to zero, we keep increasing the sum which will converge to
$$\int_{0}^{1} x^{2} dx = \frac{1}{3}.$$
This completes the proof.

We sketch a probabilistic version of the first solution. Let $x_1, x_2, x_3$, be drawn uniformly and independently at random from the segment $[0, 1]$. Let $I_1 \cup I_2 \cup \dots \cup I_n$ be a partition of $[0, 1]$ into segments of length $a_1, a_2, \dots, a_n$ in this order. Let $J_k := I_1 \cup \cdots \cup I_{k-1}$ for $k \ge 2$ and $J_1 := \emptyset$. Then
$$\begin{align*} \frac{1}{3} &= \sum_{k=1}^{n} \mathbb{P}\{x_1 \ge x_2, x_3; x_1 \in I_k\} \\ &= \sum_{k=1}^{n} \left( \mathbb{P}\{x_1 \in I_k; x_2, x_3 \in J_k\} + 2 \cdot \mathbb{P}\{x_1 \ge x_2; x_1, x_2 \in I_k; x_3 \in J_k\} \right. \\ &\qquad \left. + \mathbb{P}\{x_1 \ge x_2, x_3; x_1, x_2, x_3 \in I_k\} \right) \\ &= \sum_{k=1}^{n} \left( a_k (a_1 + \cdots + a_{k-1})^2 + 2 \cdot \frac{a_k^2}{2} \cdot (a_1 + \dots + a_{k-1}) + \frac{a_k^3}{3} \right) \\ &> \sum_{k=1}^{n} \left( a_k (a_1 + \cdots + a_{k-1})^2 + a_k^2 (a_1 + \cdots + a_{k-1}) \cdot \frac{a_1 + \cdots + a_{k-1}}{1-a_k} \right), \end{align*}$$
where for the last inequality we used that $1 - a_k \ge a_1 + \cdots + a_{k-1}$. This completes the proof since $a_k + \frac{a_k^2}{1-a_k} = \frac{a_k}{1-a_k}$.