Question 27
Let $k$ and $n$ be positive integers. Prove that, if $x_j$ are real numbers for $1 \le j \le n$, such that
$$
\sum_{j=1}^{n} \frac{1}{x_j^{2k} + k} = \frac{1}{k}
$$
holds, it follows that
$$
\sum_{j=1}^{n} \frac{1}{x_j^{2k+1} + k + 2} \le \frac{1}{k+1}
$$
$$
\sum_{j=1}^{n} \frac{1}{x_j^{2k} + k} = \frac{1}{k}
$$
holds, it follows that
$$
\sum_{j=1}^{n} \frac{1}{x_j^{2k+1} + k + 2} \le \frac{1}{k+1}
$$