Problem 1580

1. Define the sequence transformation:
Let $b_n = \frac{1}{a_n}$. Then, the given recurrence relation for $a_n$ transforms as follows:
$$a_{n+1} = \frac{a_n^2}{a_n^2 - a_n + 1}$$
implies
$$\frac{1}{a_{n+1}} = \frac{a_n^2 - a_n + 1}{a_n^2}$$
which simplifies to
$$b_{n+1} = \frac{a_n^2 - a_n + 1}{a_n^2} = 1 - \frac{a_n - 1}{a_n^2} = 1 - \frac{1}{a_n} + \frac{1}{a_n^2} = b_n^2 - b_n + 1.$$

2. Simplify the recurrence relation:
We have
$$b_{n+1} = b_n^2 - b_n + 1.$$
Consider the transformation:
$$\frac{b_{n+1} - 1}{b_n - 1} = b_n.$$

3. Telescoping product:
Multiplying these relations for $n = 1, 2, \ldots, k$, we get:
$$\prod_{i=1}^{k} \frac{b_{i+1} - 1}{b_i - 1} = \prod_{i=1}^{k} b_i.$$
This simplifies to:
$$b_{k+1} - 1 = b_1 b_2 \cdots b_k (b_1 - 1).$$

4. **Express $\frac{1}{b_n}$:**
From the above, we have:
$$b_{n+1} = b_1 b_2 \cdots b_n + 1.$$
Therefore,
$$\frac{1}{b_n} = \frac{1}{b_1 b_2 \cdots b_{n-1}} - \frac{1}{b_1 b_2 \cdots b_n}.$$

5. Telescoping sum:
Summing the above expression from $n = 1$ to $n = k$, we get:
$$\sum_{i=1}^{k} \frac{1}{b_i} = \left( \frac{1}{b_1} - \frac{1}{b_1 b_2} \right) + \left( \frac{1}{b_1 b_2} - \frac{1}{b_1 b_2 b_3} \right) + \cdots + \left( \frac{1}{b_1 b_2 \cdots b_{k-1}} - \frac{1}{b_1 b_2 \cdots b_k} \right).$$
This telescopes to:
$$\sum_{i=1}^{k} \frac{1}{b_i} = 1 - \frac{1}{b_1 b_2 \cdots b_k}.$$

6. Conclusion:
Since $b_1 = 2$ (because $a_1 = \frac{1}{2}$), we have:
$$\sum_{i=1}^{n} a_i = \sum_{i=1}^{n} \frac{1}{b_i} = 1 - \frac{1}{b_1 b_2 \cdots b_n}.$$
Since $b_i > 1$ for all $i$, it follows that $b_1 b_2 \cdots b_n > 1$, and thus:
$$\sum_{i=1}^{n} a_i < 1.$$
Therefore, we have proved that for every positive integer $n$:
$$a_1 + a_2 + \cdots + a_n < 1.$$
$\blacksquare$