17. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, satisfying $\sum_{i=1}^{n} x_{i}^{2}=1$, find the minimum value of $\frac{x_{1}^{5}}{x_{2}+x_{3}+\cdots+x_{n}}+$ $\frac{x_{2}^{5}}{x_{3}+\cdots+x_{n}+x_{1}}+\cdots+\frac{x_{n}^{5}}{x_{1}+x_{2}+\cdots+x_{n-1}}$. (2006 Turkish National Team Selection)
$$\begin{array}{l}
\text { 17. From } \sum_{i \neq j}\left(x_{i}-x_{j}\right)^{2}=(n-1) \sum_{i=1}^{n} x_{i}^{2}-2 \sum_{i \neq j} x_{i} x_{j}=(n-1)-2 \sum_{i \neq j} x_{i} x_{j} \text { we get } \\
\sum_{i=1}^{n} x_{i} \sum_{j \neq i} x_{j} \leqslant n-1 \text { and by the Cauchy-Schwarz inequality, we have }\left(\frac{x_{1}^{5}}{x_{2}+x_{3}+\cdots+x_{n}}+\frac{x_{2}^{5}}{x_{3}+\cdots+x_{n}+x_{1}}+\cdots+\right. \\
\left.\frac{-x_{n}^{5}}{x_{1}+x_{2}+\cdots+x_{n-1}}\right)\left[x_{1}\left(x_{2}+x_{3}+\cdots+x_{n}\right)+x_{2}\left(x_{3}+\cdots+x_{n}\right)+\cdots+x_{n}\left(x_{1}+\right.\right. \\
\left.\left.x_{2}+\cdots+x_{n-1}\right)\right] \geqslant\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)^{2}=n^{2}\left(\frac{x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}}{n}\right)^{2} \geqslant \\
n^{2}\left(\frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}\right)^{3}=\frac{1}{n}
\end{array}$$

The last step is obtained by the Power Mean Inequality.
Therefore,
$$\frac{x_{1}^{5}}{x_{2}+x_{3}+\cdots+x_{n}}+\frac{x_{2}^{5}}{x_{3}+\cdots x_{n}+x_{1}}+\cdots+\frac{x_{n}^{5}}{x_{1}+x_{2}+\cdots+x_{n}} \geqslant \frac{1}{n(n-1)}$$ $x_{1}+x_{2}+\cdots+x_{n-F}$ achieves the minimum value $\frac{1}{n(n-1)}$