For $n \in \mathbf{N}, a, b$ being given real numbers, and $x_{0}, x, \cdots, x_{n}$ being real variables, it is known that
$$
\sum_{i=0}^{n} x_{i}=a, \sum_{i=0}^{n} x_{i}^{2}=b .
$$

Determine the range of variation for $x_{0}$.
Since $x_{0}, x_{1}, \cdots, x_{n}$ have the same status, determining the range of $x_{0}$ is equivalent to determining the range of $x_{i}(0 \leqslant i \leqslant n)$.
Solution: By the Cauchy inequality,
$$
\left(\sum_{i=1}^{n} x_{i}\right)^{2} \leqslant n \sum_{i=1}^{n} x_{i}^{2},
$$

Therefore,
$$
\begin{aligned}
\left(a-x_{0}\right)^{2} & \leqslant n\left(b-x_{0}^{2}\right) \\
& \Rightarrow(n+1) x_{0}^{2}-2 a x_{0}+a^{2}-n b \leqslant 0 .
\end{aligned}
$$

The discriminant of this quadratic trinomial is $D=4 n(n+1)\left(b-\frac{a^{2}}{n+1}\right)$, hence
(1) If $b\frac{a^{2}}{n+1}$, then $D>0$, at this time
$$
\frac{a-\sqrt{\frac{D}{4}}}{n+1} \leqslant x_{0} \leqslant \frac{a+\sqrt{\frac{D}{4}}}{n+1} .
$$

When taking $a=p, b=\frac{p^{2}}{n}$, obviously, $b=\frac{p^{2}}{n}>\frac{p^{2}}{n+1}=\frac{a^{2}}{n+1}$, hence by (3) we get
$$
\begin{array}{c}
D=4 n(n+1)\left(\frac{p^{2}}{n}-\frac{p^{2}}{n+1}\right)=4 p^{2}, \\
\frac{p-\sqrt{\frac{4 p^{2}}{4}}}{n} \leqslant x_{0} \leqslant \frac{p+\sqrt{\frac{4 p^{2}}{4}}}{n},
\end{array}
$$

Thus,
$$
0 \leqslant x_{0} \leqslant \frac{2 p}{n} \text {. }
$$

Proof completed.