Problem 1701

Given the sequences of points $A_n$ on the positive half of the y-axis and $B_n$ on the curve $y = \sqrt{2x}$ with the condition $|OA_n| = |OB_n| = \frac{1}{n}$, we need to prove two statements.

1. $a_n > a_{n+1} > 4$ for all $n \in \mathbb{N}$.
2. There exists $n_0 \in \mathbb{N}$ such that for any $n > n_0$, $\frac{b_2}{b_1} + \frac{b_3}{b_2} + \ldots + \frac{b_{n+1}}{b_n} < n - 2004$.

### Part 1: Proving $a_n > a_{n+1} > 4$

1. **Determine Coordinates of $A_n$ and $B_n$**:
- $A_n = \left(0, \frac{1}{n}\right)$
- $B_n = (b_n, \sqrt{2b_n})$
- Given $|OB_n| = \frac{1}{n}$, we have:
$$b_n^2 + 2b_n = \frac{1}{n^2} \implies b_n = \sqrt{1 + \frac{1}{n^2}} - 1$$

2. **Find the x-intercept $a_n$ of the line $A_nB_n$**:
- The slope of the line $A_nB_n$ is:
$$\text{slope} = \frac{\sqrt{2b_n} - \frac{1}{n}}{b_n - 0} = \frac{\sqrt{2b_n} - \frac{1}{n}}{b_n}$$
- The equation of the line is:
$$y - \frac{1}{n} = \left(\frac{\sqrt{2b_n} - \frac{1}{n}}{b_n}\right)x$$
- Setting $y = 0$ to find the x-intercept $a_n$:
$$0 - \frac{1}{n} = \left(\frac{\sqrt{2b_n} - \frac{1}{n}}{b_n}\right)a_n \implies -\frac{1}{n} = \left(\frac{\sqrt{2b_n} - \frac{1}{n}}{b_n}\right)a_n$$
$$a_n = \frac{b_n}{\sqrt{2b_n} - \frac{1}{n}} = \frac{b_n(1 + n\sqrt{2b_n})}{1 - 2n^2b_n} = \frac{1 + n\sqrt{2b_n}}{n^2b_n}$$

3. **Simplify $a_n$**:
$$a_n = \frac{1}{n^2b_n} + \sqrt{\frac{2}{n^2b_n}} = b_n + 2 + \sqrt{2(b_n + 2)}$$

4. **Show $a_n > a_{n+1}$**:
- Since $\frac{1}{n^2} > \frac{1}{(n+1)^2}$, we have $b_n > b_{n+1}$.
- Therefore:
$$a_n = b_n + 2 + \sqrt{2(b_n + 2)} > b_{n+1} + 2 + \sqrt{2(b_{n+1} + 2)} = a_{n+1}$$

5. **Show $a_n > 4$**:
- Since $b_n > 0$ for all $n \in \mathbb{N}$:
$$a_n = b_n + 2 + \sqrt{2(b_n + 2)} > 4$$

### Part 2: Proving the Inequality for $b_n$

1. **Define $c_n$**:
$$c_n = 1 - \frac{b_{n+1}}{b_n}$$

2. **Simplify $c_n$**:
$$c_n = 1 - \frac{\sqrt{1 + \frac{1}{(n+1)^2}} - 1}{\sqrt{1 + \frac{1}{n^2}} - 1}$$
$$c_n = \frac{\sqrt{1 + \frac{1}{n^2}} - \sqrt{1 + \frac{1}{(n+1)^2}}}{\sqrt{1 + \frac{1}{n^2}} - 1}$$
$$c_n = n^2\left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) \cdot \frac{\sqrt{1 + \frac{1}{n^2}} + 1}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 + \frac{1}{(n+1)^2}}}$$
$$c_n > \frac{2n + 1}{(n + 1)^2}\left(\frac{1}{2} + \frac{1}{2\sqrt{\frac{1}{n^2} + 1}}\right)$$
$$c_n > \frac{2n + 1}{2(n + 1)^2} = \frac{2n + 1}{(2n + 1)(n + 2) - n} > \frac{1}{n + 2}$$

3. **Sum $c_n$**:
$$S_n = c_1 + c_2 + \dots + c_n$$
- Let $n = 2^k - 2$ for some positive integer $k \geq 2$:
$$S_n > \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^k - 1} + \frac{1}{2^k}$$
$$= \left(\frac{1}{2 + 1} + \frac{1}{2^2}\right) + \left(\frac{1}{2^2 + 1} + \frac{1}{2^3}\right) + \dots + \left(\frac{1}{2^{k-1} + 1} + \frac{1}{2^k}\right)$$
$$> 2 \cdot \frac{1}{2^2} + 2^2 \cdot \frac{1}{2^3} + \dots + 2^{k-1} \cdot \frac{1}{2^k} = \frac{k - 1}{2}$$

4. **Set $n_0$**:
- If we set $n_0 = 2^{4009} - 2$, then for all $n > n_0$:
$$\left(1 - \frac{b_2}{b_1}\right) + \left(1 - \frac{b_3}{b_2}\right) + \dots + \left(1 - \frac{b_{n+1}}{b_n}\right) = S_n > S_{n_0} > \frac{4009-1}{2} = 2004$$
$$\frac{b_2}{b_1} + \frac{b_3}{b_2} + \dots + \frac{b_n}{b_{n-1}} + \frac{b_{n+1}}{b_n} < n - 2004$$