3 Given a natural number $n>2, \lambda$ is a given constant. Determine the maximum and minimum values of the function: $F=x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{n}^{2}+\lambda x_{1} x_{2} \cdots x_{n}$, where $x_{1}, x_{2}, \cdots, x_{n}$ are non-negative real numbers, and $x_{1}+x_{2}+\cdots+x_{n}=1$. (1994 China Mathematics Training Team Test Question)
3. Since $F$ is continuous on the closed domain $0 \leqslant x_{i} \leqslant 1$, $F$ must have a maximum and a minimum value. Suppose $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is a point where $F$ attains its extremum. We will prove that for any $i \neq j, x_{i} x_{j}=0$, or $x_{i}=x_{j}$. In fact, by symmetry, we only need to consider $x_{1}$ and $x_{2}$. Fix $x_{3}, x_{4}, \cdots, x_{n}$, then $x_{1}+x_{2}=1-\left(x_{3}+x_{4}+\cdots+x_{n}\right)=c$ (a constant). $F=\left(x_{1}+x_{2}\right)^{2}+x_{3}^{2}+\cdots+x_{n}^{2}-2 x_{1} x_{2}+\lambda x_{1} x_{2} \cdots x_{n}=\left(x_{1}+x_{2}\right)^{2}+x_{3}^{2}+\cdots+x_{n}^{2}+\left(\lambda x_{3} x_{4} \cdots x_{n}-2\right) \cdot x_{1}\left(c-x_{1}\right)$. Let $f\left(x_{1}\right)=x_{1}\left(c-x_{1}\right)$, since $0 \leqslant x_{1} \leqslant c$, by the properties of quadratic functions, when $f\left(x_{1}\right)$ attains its extremum, $x_{1} \in\left\{0, c, \frac{c}{2}\right\}$. Note that at this time $x_{1}+x_{2}=c$, thus $x_{1}$ and $x_{2}$ have the following relationships: (1) $x_{1}=0, x_{2}=c$. (2) $x_{1}=c, x_{2}=0$. (3) $x_{1}=x_{2}=\frac{c}{2}$. Therefore, either $x_{1} x_{2}=0$, or $x_{1}=x_{2}$. From the above discussion, if $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is an extremum point of $F$, then the non-zero numbers among $x_{1}, x_{2}, \cdots, x_{n}$ are all equal. Suppose $x_{1}=x_{2}=\cdots=x_{k} \neq 0$, $x_{k+1}+x_{k+2}+\cdots+x_{n}=0$, there are the following cases: (1) If $k=n$, then $F$ attains its extremum at $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n}$, at this time $F=\sum_{i=1}^{n} \frac{1}{n^{2}}+\lambda \prod_{i=1}^{n} \frac{1}{n}=\frac{\lambda+n^{n-1}}{n^{n}}$. (2) If $k<n$, then $F$ attains its extremum at $x_{1}=x_{2}=\cdots=x_{k}=\frac{1}{k}, x_{k+1}=x_{k+2}=\cdots=x_{n}=0$, at this time $F=\sum_{i=1}^{k} \frac{1}{k^{2}}=\frac{1}{k}$. Note that $1 \leqslant k \leqslant n-1$, so $\frac{1}{n-1} \leqslant F \leqslant 1$. Therefore, the set of extremum values of $F$ is $\left\{1, \frac{1}{n-1}, \frac{\lambda+n^{n-1}}{n^{n}}\right\}$. Hence $F_{\min }=\min \left\{\frac{1}{n-1}, \frac{\lambda+n^{n-1}}{n^{n}}\right\}, F_{\max }=\max \left\{1, \frac{\lambda+n^{n-1}}{n^{n}}\right\}$.