Problem 1814

To address the problem, we need to prove that there exists a constant $B > 0$ such that $a_n \leq \frac{B}{n}$ for all $n \geq 1$. Let's break down the solution step-by-step and address the two specific questions.

1. **Why it suffices to prove it for $n = 2^k$:**

The idea is to use a subsequence of the original sequence to simplify the problem. If we can show that $a_{2^k} \leq \frac{B}{2^k}$ for all $k$, then we can extend this result to all $n$ by using the fact that the sequence $(a_n)$ is monotonically decreasing.

Specifically, if $a_{2^k} \leq \frac{B}{2^k}$ for all $k$, then for any $n$ such that $2^k \leq n < 2^{k+1}$, we have:
$$a_n \leq a_{2^k} \leq \frac{B}{2^k} \leq \frac{B}{n}$$
because $2^k \leq n$ implies $\frac{1}{2^k} \leq \frac{1}{n}$.

2. **How can we assume $A = 2$:**

The assumption $A = 2$ is a rescaling argument. We can rescale the sequence $(a_n)$ by a constant factor to simplify the inequality. Specifically, if we let $A$ be any positive constant, we can define a new sequence $(b_n)$ such that $b_n = \sqrt{A} a_n$. Then the inequality $a_m - a_{m+1} \geq A a_m^2$ transforms into:
$$b_m - b_{m+1} \geq 2 b_m^2$$
by choosing the constant factor appropriately. This allows us to work with the simpler case $A = 2$.

Now, let's proceed with the detailed solution:

1. Monotonicity and initial setup:

Given $a_n \geq 0$ and $a_m - a_{m+1} \geq A a_m^2$, we know that the sequence $(a_n)$ is monotonically decreasing. We aim to show that $a_n \leq \frac{B}{n}$ for some constant $B > 0$.

2. **Induction and bounding $a_{2^k}$:**

We use induction to show that $a_{2^k} \leq \frac{B}{2^k}$ for all $k$. Assume $A = 2$ for simplicity. Define $b_n = 2^n a_{2^n}$. We need to show that $b_n$ is bounded.

3. Inequality transformation:

From the given inequality, we have:
$$a_{m+d} \leq a_m - A \sum_{i=0}^{d-1} a_{m+i}^2$$
For $d = m$, this becomes:
$$a_{2m} \leq a_m - A \sum_{i=0}^{m-1} a_{m+i}^2$$
Using the rescaled sequence $b_n = 2^n a_{2^n}$, we get:
$$b_{n+1}(1 + b_{n+1}) \leq 2 b_n$$

4. **Bounding $b_n$:**

We analyze the behavior of $b_n$:
- If $b_{n+1} \geq 1$, then $b_{n+1} \leq b_n$.
- If $b_{n+1} \leq 1$, then $b_{n+1}^2 \leq b_n$.

In both cases, $b_n$ is non-increasing and hence bounded.

5. Conclusion:

Since $b_n$ is bounded, there exists a constant $C$ such that $b_n \leq C$ for all $n$. Therefore, $2^n a_{2^n} \leq C$, which implies:
$$a_{2^n} \leq \frac{C}{2^n}$$
For any $n$, we have $2^k \leq n < 2^{k+1}$ for some $k$, and thus:
$$a_n \leq a_{2^k} \leq \frac{C}{2^k} \leq \frac{C}{n}$$
Hence, we can choose $B = C$ to satisfy $a_n \leq \frac{B}{n}$ for all $n$.

The final answer is $\boxed{a_n \leq \frac{B}{n}}$