Problem 1682

1. Define the sequence and initial conditions:
Let $S_n = \sum_{i=1}^n a_i$. Given the condition $a_1 + a_2 + \cdots + a_{m+1} \leq r a_m$, we can rewrite it as $S_{m+1} \leq r a_m$.

2. **Establish the increasing nature of $S_n$:**
Since $a_n > 0$ for all $n$, the sequence $S_n$ is strictly increasing. This implies $S_{n+1} > S_n$, and thus $r > 1$.

3. **Define the set $K$ and its supremum $\ell$:**
Define $K = \{ k \in \mathbb{R}_+^* \mid S_{n+1} \geq k S_n \text{ for all } n \in \mathbb{Z}_+^* \}$. The set $K$ is non-empty because $1 \in K$ (since $S_{n+1} > S_n$). Also, $K$ is bounded above by $\frac{S_2}{S_1}$. Let $\ell = \sup K$.

4. **Show that $\ell \in K$:**
Since $\ell$ is the supremum of $K$, for any $\epsilon > 0$, there exists $k \in K$ such that $\ell - \epsilon < k \leq \ell$. This implies $S_{n+1} \geq k S_n$ for all $n$, and by taking the limit as $\epsilon \to 0$, we get $S_{n+1} \geq \ell S_n$.

5. **Derive the inequality involving $r$ and $\ell$:**
From the given condition $S_{m+1} \leq r a_m$, we have $S_{n+2} \leq r (S_{n+1} - S_n)$. Using $S_{n+1} \geq \ell S_n$, we get:
$$\ell S_{n+1} \leq S_{n+2} \leq r (S_{n+1} - S_n)$$
This simplifies to:
$$\ell S_{n+1} \leq r S_{n+1} - r S_n$$
Dividing both sides by $S_{n+1}$ and using $S_{n+1} \geq \ell S_n$, we get:
$$\ell \leq r - \frac{r S_n}{S_{n+1}} \leq r - \frac{r}{\ell}$$
Therefore:
$$\ell \leq r - \frac{r}{\ell}$$
Multiplying both sides by $\ell$, we obtain:
$$\ell^2 \leq r \ell - r$$
Rearranging terms, we get the quadratic inequality:
$$\ell^2 - r \ell + r \leq 0$$

6. Analyze the discriminant of the quadratic inequality:
For the quadratic inequality $\ell^2 - r \ell + r \leq 0$ to have real solutions, its discriminant must be non-negative:
$$\Delta = r^2 - 4r \geq 0$$
Solving for $r$, we get:
$$r^2 - 4r \geq 0 \implies r(r - 4) \geq 0$$
This implies $r \geq 4$ or $r \leq 0$. Since $r > 1$, we conclude $r \geq 4$.

7. **Verify that $r = 4$ is achievable:**
Consider the sequence defined by $a_1 = a$ and $a_n = 2^{n-2}a$ for $n \geq 2$. Then:
$$S_{n+1} = 2 S_n \implies S_n = 2^{n-1}a$$
This gives $\ell = 2$ and $r = 4$, satisfying the given condition.

$\blacksquare$

The final answer is $\boxed{ r \geq 4 }$