26. (KOR 4) Let $n$ be a positive integer and let $a, b$ be given real numbers. Determine the range of $x_{0}$ for which  
$$ \sum_{i=0}^{n} x_{i}=a \quad \text { and } \quad \sum_{i=0}^{n} x_{i}^{2}=b $$  
where $x_{0}, x_{1}, \ldots, x_{n}$ are real variables.
26. By the Cauchy-Schwarz inequality, 
$$ \left(\sum_{i=1}^{n} x_{i}\right)^{2} \leq n \sum_{i=1}^{n} x_{i}^{2} $$
Since $\sum_{i=1}^{n} x_{i}=a-x_{0}$ and $\sum_{i=1}^{n} x_{i}^{2}=b-x_{0}^{2}$, we have $\left(a-x_{0}\right)^{2} \leq$ 
$n\left(b-x_{0}^{2}\right)$, i.e., 
$$ (n+1) x_{0}^{2}-2 a x_{0}+\left(a^{2}-n b\right) \leq 0 . $$
The discriminant of this quadratic is $D=4 n(n+1)\left[b-a^{2} /(n+1)\right]$, so we conclude that (i) if $a^{2}>(n+1) b$, then such an $x_{0}$ does not exist; (ii) if $a^{2}=(n+1) b$, then $x_{0}=a / n+1 ; \quad$ and (iii) if $a^{2}<(n+1) b$, then $\frac{a-\sqrt{D} / 2}{n+1} \leq x_{0} \leq \frac{a+\sqrt{D} / 2}{n+1}$. 
It is easy to see that these conditions for $x_{0}$ are also sufficient.