Let $ n \in \mathbb{Z}^+$ and let $ a, b \in \mathbb{R}.$ Determine the range of $ x_0$ for which
\[ \sum^n_{i=0} x_i = a \text{ and } \sum^n_{i=0} x^2_i = b,\]
where $ x_0, x_1, \ldots , x_n$ are real variables.
The Power-Mean Inequality dictates that \[\frac{b-x_0^2}{n} = \frac{1}{n}\sum_{i=1}^n x_i^2 \geq \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2 = \left(\frac{a-x_0}{n}\right)^2\,.\] Therefore, \[(n+1)x_0^2-2ax_0+\left(a^2-nb\right) \leq 0\,.\] Let $D:=\sqrt{n\left((n+1)b-a^2\right)}$ (and so, we implicitly require $a^2 \leq (n+1)b$, another instance of the Power-Mean Ineq.). Ergo, the range of $x_0$ is a subset of $\left[\frac{a-D}{n+1},\frac{a+D}{n+1}\right]$. It is easy to see that this range is exactly the answer.