Problem 1791

To prove the inequality
$$\sum_{i = 1}^n \frac{a_i}{\sqrt{a_1^2 + \dots + a_i^2}} \le 4 \sqrt{kn},$$
we will proceed with the following steps:

1. Initial Setup and Assumptions:
Let $f(a_1, \dots, a_n) = \sum_{i=1}^n \frac{a_i}{\sqrt{a_1^2 + \dots + a_i^2}}$. We claim that if $a_i > a_{i+1}$, then swapping $a_i$ and $a_{i+1}$ will increase the value of $f$.

2. Proof of the Claim:
Let $m = a_1^2 + \dots + a_{i-1}^2$. We need to show that:
$$\frac{a_{i+1}}{\sqrt{m + a_{i+1}^2}} + \frac{a_i}{\sqrt{m + a_{i+1}^2 + a_i^2}} > \frac{a_i}{\sqrt{m + a_i^2}} + \frac{a_{i+1}}{\sqrt{m + a_i^2 + a_{i+1}^2}}.$$
This can be rewritten as:
$$a_{i+1} \left( \frac{1}{\sqrt{m + a_{i+1}^2}} - \frac{1}{\sqrt{m + a_i^2 + a_{i+1}^2}} \right) > a_i \left( \frac{1}{\sqrt{m + a_i^2}} - \frac{1}{\sqrt{m + a_i^2 + a_{i+1}^2}} \right).$$
Simplifying further, we get:
$$\frac{a_{i+1} a_i^2}{\sqrt{m + a_{i+1}^2} (\sqrt{m + a_{i+1}^2} + \sqrt{m + a_i^2 + a_{i+1}^2})} > \frac{a_i a_{i+1}^2}{\sqrt{m + a_i^2} (\sqrt{m + a_i^2} + \sqrt{m + a_i^2 + a_{i+1}^2})}.$$
This inequality holds because $a_i > a_{i+1}$. Hence, the claim is proved. $\blacksquare$

3. Reduction to Ordered Sequence:
By the above claim, it suffices to consider the case when $1 = a_1 \leq a_2 \leq \dots \leq a_n = 2^k$.

4. Transformation to Trigonometric Form:
Let $a_2 = \tan \theta_2$ and
$$a_i = \left( \prod_{j=1}^{i-1} \sec \theta_j \right) \cdot \tan \theta_i.$$
The conditions translate to:
$$\begin{cases} \tan \theta_2 \geq 1, \\ \tan \theta_3 \geq \sin \theta_2, \\ \dots \\ \tan \theta_n \geq \sin \theta_{n-1}, \\ \sec \theta_2 \sec \theta_3 \dots \sec \theta_{n-1} \tan \theta_n \leq 2^k. \end{cases}$$

5. Simplification Using Trigonometric Identities:
We use the sum-to-product and product-to-sum identities to show that if $\theta_i > \theta_j$ and we replace $(\theta_i, \theta_j)$ with $(\theta_i - x, \theta_j + x)$ such that $\theta_i - x > \theta_j + x$, the value of $f$ increases.

6. Final Bound Calculation:
We now mix the variables without violating the conditions. After finitely many refinements, we achieve the state where the sequence is of the form:
$$\underbrace{1, \dots, 1}_{i \text{ times}}, a, a \left( \frac{a^2}{i} + 1 \right)^{\frac{1}{2}}, a \left( \frac{a^2}{i} + 1 \right), \dots, a \left( \frac{a^2}{i} + 1 \right)^{\frac{m}{2}}, \underbrace{2^k, \dots, 2^k}_{j \text{ times}}.$$

7. Integral Approximation:
Using integral approximation, we get:
$$\sum_{p=1}^i \frac{a_p}{\sqrt{a_1^2 + \dots + a_p^2}} + \sum_{p=i+m+2}^{i+m+1+j} \frac{a_p}{\sqrt{a_1^2 + \dots + a_p^2}} < \int_1^i \frac{1}{\sqrt{x}} \, dx + \int_2^{j+1} \frac{1}{\sqrt{x}} \, dx \leq 2 \sqrt{i} + 2 \sqrt{j} - 2 < \sqrt{8(i+j)} - 2.$$

8. Combining Bounds:
Combining all these bounds, we have:
$$\sum_{i=1}^n \frac{a_i}{\sqrt{a_1^2 + \dots + a_i^2}} \leq \sqrt{8(i+j)} + \sqrt{2 \ln 2 km} \leq \sqrt{16(i+j) + 4 \ln 2 km} \leq 4 \sqrt{k(i+j+m+1)} = 4 \sqrt{kn}.$$

Thus, the inequality is proved.