To prove the inequality
$$a_1\left(b_1 + a_2\right) + a_2\left(b_2 + a_3\right) + \cdots + a_{n-1}\left(b_{n-1} + a_n\right) + a_n\left(b_n + a_1\right) < 1,$$
we will use the given conditions:
$$a_1 + a_2 + \cdots + a_n = 1 \quad \text{and} \quad b_1^2 + b_2^2 + \cdots + b_n^2 = 1.$$

1. Express the sum in terms of two separate sums:
$$\sum_{i=1}^n a_i(b_i + a_{i+1}) = \sum_{i=1}^n a_i b_i + \sum_{i=1}^n a_i a_{i+1},$$
where $a_{n+1} = a_1$.

2. **Apply the Cauchy-Schwarz inequality to $\sum_{i=1}^n a_i b_i$:**
$$\left( \sum_{i=1}^n a_i b_i \right)^2 \leq \left( \sum_{i=1}^n a_i^2 \right) \left( \sum_{i=1}^n b_i^2 \right).$$
Given $\sum_{i=1}^n b_i^2 = 1$, we have:
$$\left( \sum_{i=1}^n a_i b_i \right)^2 \leq \sum_{i=1}^n a_i^2.$$
Therefore,
$$\sum_{i=1}^n a_i b_i \leq \sqrt{\sum_{i=1}^n a_i^2}.$$

3. **Use the fact that $\sum_{i=1}^n a_i = 1$ to bound $\sum_{i=1}^n a_i^2$:**
By the QM-AM inequality (Quadratic Mean - Arithmetic Mean inequality),
$$\sqrt{\frac{\sum_{i=1}^n a_i^2}{n}} \geq \frac{\sum_{i=1}^n a_i}{n} = \frac{1}{n}.$$
Thus,
$$\sum_{i=1}^n a_i^2 \geq \frac{1}{n}.$$

4. **Consider the sum $\sum_{i=1}^n a_i a_{i+1}$:**
By the AM-GM inequality (Arithmetic Mean - Geometric Mean inequality),
$$a_i a_{i+1} \leq \frac{a_i^2 + a_{i+1}^2}{2}.$$
Summing over all $i$,
$$\sum_{i=1}^n a_i a_{i+1} \leq \frac{1}{2} \sum_{i=1}^n (a_i^2 + a_{i+1}^2) = \sum_{i=1}^n a_i^2.$$

5. Combine the results:
$$\sum_{i=1}^n a_i b_i + \sum_{i=1}^n a_i a_{i+1} \leq \sqrt{\sum_{i=1}^n a_i^2} + \sum_{i=1}^n a_i^2.$$
Since $\sum_{i=1}^n a_i^2 \leq 1$ and $\sqrt{\sum_{i=1}^n a_i^2} \leq 1$,
$$\sqrt{\sum_{i=1}^n a_i^2} + \sum_{i=1}^n a_i^2 < 1.$$

Therefore, we have shown that:
$$a_1\left(b_1 + a_2\right) + a_2\left(b_2 + a_3\right) + \cdots + a_{n-1}\left(b_{n-1} + a_n\right) + a_n\left(b_n + a_1\right) < 1.$$

$\blacksquare$