SOLUTION: Let x_1, x_2, \dots, x_n be real numbers. If x_1^2 + 2x_2^2 + \dots + nx_n^2 = 1, then find the maximum value of (x_1 + x_2/2 + \dots + x

SOLUTION: Let x_1, x_2, \dots, x_n be real numbers. If x_1^2 + 2x_2^2 + \dots + nx_n^2 = 1, then find the maximum value of (x_1 + x_2/2 + \dots + x_n/n)^2, in terms of n.

Question 1209919: Let x_1, x_2, \dots, x_n be real numbers. If
x_1^2 + 2x_2^2 + \dots + nx_n^2 = 1,
then find the maximum value of (x_1 + x_2/2 + \dots + x_n/n)^2, in terms of n.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let $a_k = x_k \sqrt{k}$ for $k = 1, 2, \dots, n$.
Then the given condition becomes
$$\sum_{k=1}^n x_k^2 k = \sum_{k=1}^n a_k^2 = 1$$
We want to find the maximum value of
$$S = \left( \sum_{k=1}^n \frac{x_k}{k} \right)^2 = \left( \sum_{k=1}^n \frac{a_k}{k\sqrt{k}} \right)^2$$
By the Cauchy-Schwarz inequality, we have
$$\left( \sum_{k=1}^n \frac{a_k}{k\sqrt{k}} \right)^2 \le \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n \frac{1}{k^3} \right)$$
Since $\sum_{k=1}^n a_k^2 = 1$, we have
$$S \le \sum_{k=1}^n \frac{1}{k^3}$$
The maximum value of $\left( \sum_{k=1}^n \frac{x_k}{k} \right)^2$ is $\sum_{k=1}^n \frac{1}{k^3}$.
To achieve equality in Cauchy-Schwarz inequality, we need
$$\frac{a_1}{1\sqrt{1}} = \frac{a_2}{2\sqrt{2}} = \dots = \frac{a_n}{n\sqrt{n}} = c$$
where $c$ is a constant.
Then $a_k = ck\sqrt{k}$, so $x_k \sqrt{k} = ck\sqrt{k}$, which means $x_k = ck$.
Substituting into the given condition, we have
$$\sum_{k=1}^n kx_k^2 = \sum_{k=1}^n k(ck)^2 = c^2 \sum_{k=1}^n k^3 = 1$$
$$c^2 = \frac{1}{\sum_{k=1}^n k^3}$$
$$c = \frac{1}{\sqrt{\sum_{k=1}^n k^3}}$$
Then $x_k = \frac{k}{\sqrt{\sum_{k=1}^n k^3}}$.
Therefore, the maximum value of $\left( \sum_{k=1}^n \frac{x_k}{k} \right)^2$ is $\sum_{k=1}^n \frac{1}{k^3}$.
Final Answer: The final answer is $\boxed{\sum_{k=1}^n \frac{1}{k^3}}$

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