orl wrote: Let $ a_1, a_2, \ldots, a_{2n}$ be $ 2n$ real numbers satisfying the condition $ \sum_{i = 1}^{2n - 1} (a_{i + 1} - a_i)^2 = 1$. Find the greatest possible value of $ (a_{n + 1} + a_{n + 2} + \ldots + a_{2n}) - (a_1 + a_2 + \ldots + a_n)$. I don't think this problem should be in this section. Anyway, here is a brief hint.

The equality you wrote in the hint doesn't agree about the coefficient of $ a_1$. In LH its -1, in in RH it's -(2n-1). I think you meant min(k,2n-k). Perhaps I'm wrong... My solution: Denote $ x_n = a_{n + 1} - a_n$, and vector $ x = (x_n)$ The condition is $ ||x||^2 = 1$ And the expression to maximize can be written as $ x\cdot(1,2,\cdots,n - 1,n,n - 1,\cdots,2,1)$ So, we need to take $ a = (1,2,\cdots,n - 1,n,n - 1,\cdots,2,1)$ and normalize it to 1: After using sum of squares formula, $ ||a||^2 = \frac {n(n + 1)(2n + 1) + (n - 1)n(2n - 1)}{6} = \frac {n(2n^2 + 1)}{3}$ So, our vector will be $ x = \frac {a}{||a||}$, and $ x\cdot a = ||a||$ I hope I didn't miscalculate anything important, but I think the idea is right.

Putting $ x_1 = a_1 ; x_2 = a_2 - a_1 ; x_3 = a_3-a_2 ; ... ; x_{2n}= a_{2n} - a_{2n-1}$ We have $ x_2^2 + x_3^2 + .. + x_{2n}^2 = 1$ $ a_1 = x_1 ; a_2 = x_1 + x_2 ; a_3 = x_1 + x_2 + x_3 ; .. ; a_{2n} = x_1 + x_2 +..+x_{2n}$ So $ S=(a_{n+1} + a_{n+2} +..+ a_{2n} ) - ( a_1 + a_2 + ..+ a_n ) = x_2 + 2x_3 + ... + (n-1)x_n + nx_{n+1} + (n-1)x_{n+2} + .. + x_{2n}$ Applying Cauchy-Swartz inequality : $ S^2 \leq ( x_2^2 + x_3^2 + ...+ x_{2n}^2 )(1^2 + 2^2 + ... + n^2 + ... + 1^2 )$ It's easy to check that the equality does hold

Synthesize a sequence $\{b_n\}$ by $a_1 = b_1$ and $b_i = a_i - a_{i-1}$ for $i \geq 2$. Let $S$ be the desired expression. We have $$\left(\sum b_i^2\right)(1^2+2^2+3^2+\cdots+n^2+(n-1)^2+\cdots+1^2) \geq S^2$$by Cauchy-Schwarz, so $$S^2 \leq \frac 13(2n^3+n) \iff S \leq \sqrt{\frac 13(2n^3+n)}.$$Equality obviously holds.