jred wrote: Let $x_1,x_2,\ldots ,x_n$ be real numbers, where $n\ge 2$, satisfying $\sum_{i=1}^{n}x^2_i+ \sum_{i=1}^{n-1}x_ix_{i+1}=1$ . For each $k$, find the maximal value of $|x_k|$.

I think the answer is (2k(n-k+1)/(n+1))^(1/2)

Supravat wrote: I think the answer is (2k(n-k+1)/(n+1))^(1/2) How do you prove this $?$

$1=\displaystyle\sum_{i=1}^n{x_i^2}+\sum_{i=1}^{n-1}{x_ix_{i+1}}\\=\displaystyle\sum_{i=1}^k{\frac{i+1}{2i}(x_i+\frac{i}{i+1}x_{i+1})^2}+\sum_{i=1}^{n-k}{\frac{i+1}{2i}(x_{n+1-i}+\frac{i}{i+1}x_{n-i})^2}+\frac{n+1}{2k(n-k+1)}x_k^2\\\geq\displaystyle\frac{n+1}{2k(n-k+1)}x_k^2$ $\therefore |x_k|\leq\displaystyle\sqrt{\frac{2k(n-k+1)}{n+1}}$