For example, $4 n \in N, a, b \in R, x_{a}, x_{1}, \cdots, x_{n}$ are real variables, and $\sum_{i=0}^{n} x_{i}=a, \sum_{i=1}^{n} x_{i}^{2}=b$. Try to determine the range of $x_{0}$.
(30th IMO Preliminary Problem)
Let $x_{i}=\frac{a-x_{0}}{n}+t_{i}, i=1,2, \cdots, n$. Where $t_{1}+t_{2}+\cdots+t_{n}=0$. Then
$$
\begin{aligned}
x_{0}^{2}-b= & -\sum_{i=1}^{n} x_{i}^{2}=-\sum_{i=1}^{n}\left(\frac{a-x_{0}}{n}+t_{i}\right)^{2} \\
= & -\sum_{i=1}^{n}\left(\frac{a-x_{0}}{n}\right)^{2} \\
& -2 \cdot \frac{a-x_{0}}{n} \cdot \sum_{i=1}^{n} t_{i}-\sum_{i=1}^{n} t_{i}^{2} \\
= & -\sum_{i=1}^{n}\left(\frac{a-x_{0}}{n}\right)^{2}-\sum_{i=1}^{n} t_{i}^{2} \\
\leqslant & -\frac{(a-x_{0})^{2}}{n},
\end{aligned}
$$

i.e., $n\left(x_{0}^{2}-b\right)+\left(a-x_{0}\right)^{2} \leqslant 0$.
$$
\text { Solving it, we get } \begin{aligned}
& \frac{a-\sqrt{n^{2} b-n a^{2}+n b}}{n+1} \\
& \leqslant x_{0} \leqslant \frac{a+\sqrt{n^{2} b-n a^{2}+n b}}{n+1} .
\end{aligned}
$$