The fifteenth question: Given a positive integer $n \geq 2$, find the largest positive real number $\lambda(n)$, such that for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, not all positive, we have $\prod_{i=1}^{n}\left(x_{i}^{2}-x_{i}+1\right) \geq \lambda(n) \cdot\left(\left(\prod_{i=1}^{n} x_{i}\right)^{2}-\prod_{i=1}^{n} x_{i}+1\right)$.
Fifteenth question,
Solution: On one hand, take $x_{1}=0, x_{2}=x_{3}=\ldots=x_{n}=\frac{1}{2}$, then $\lambda(n) \leq\left(\frac{3}{4}\right)^{n-1}$. Below, we prove: if $x_{1}, x_{2}, \cdots, x_{n}$ are not all positive real numbers, then
$$
\prod_{i=1}^{n}\left(x_{i}^{2}-x_{i}+1\right) \geq\left(\frac{3}{4}\right)^{n-1} \cdot\left(\left(\prod_{i=1}^{n} x_{i}\right)^{2}-\prod_{i=1}^{n} x_{i}+1\right)
$$

We use induction on $n$.
When $n=2$, without loss of generality, assume $x_{1} \leq 0$, and consider two cases.
(1) If $x_{2} \leq 0$, then by the Cauchy-Schwarz inequality, we have:
$$
\begin{array}{l}
\left(x_{1}^{2}-x_{1}+1\right)\left(x_{2}^{2}-x_{2}+1\right)=\left(x_{1}^{2}+\left|x_{1}\right|+1\right)\left(x_{2}^{2}+\left|x_{2}\right|+1\right) \\
\geq\left(\left|x_{1} x_{2}\right|+\sqrt{\left|x_{1} x_{2}\right|}+1\right)^{2} \geq\left(x_{1} x_{2}\right)^{2}+\left|x_{1} x_{2}\right|+1 \geq \frac{3}{4}\left(\left(x_{1} x_{2}\right)^{2}-x_{1} x_{2}+1\right)
\end{array}
$$
(2) If $x_{2}>0$, then by completing the square, we have:
$$
\begin{array}{l}
\left(x_{1}^{2}-x_{1}+1\right)\left(x_{2}^{2}-x_{2}+1\right)-\frac{3}{4}\left(\left(x_{1} x_{2}\right)^{2}-x_{1} x_{2}+1\right) \\
=\frac{4\left(x_{1}^{2}+\left|x_{1}\right|+1\right)\left(x_{2}^{2}-\left|x_{2}\right|+1\right)-3\left(\left(x_{1} x_{2}\right)^{2}+\left|x_{1} x_{2}\right|+1\right)}{4} \\
=\frac{x_{1}^{2}\left(\left|x_{2}\right|-2\right)^{2}+\left(2\left|x_{2}\right|-1\right)^{2}+4\left|x_{1}\right|\left(\left|x_{2}\right|-1\right)^{2}+\left|x_{1} x_{2}\right|}{4} \geq 0
\end{array}
$$

The conclusion holds.
Assume the conclusion holds for all positive integers less than $n (n \geq 3)$. Now consider the case for $n$.
Without loss of generality, assume $x_{n} \leq 0$. Consider two cases.
(1) If $x_{1}, x_{2}, \ldots, x_{n-1}$ contain a number greater than 0, without loss of generality, assume $x_{n-1}>0$, then $x_{n-1} x_{n} \leq 0$. By the induction hypothesis, we have
$$
\begin{array}{l}
\prod_{i=1}^{n}\left(x_{i}^{2}-x_{i}+1\right) \geq \prod_{i=1}^{n-2}\left(x_{i}^{2}-x_{i}+1\right) \cdot \frac{3}{4}\left(\left(x_{n-1} x_{n}\right)^{2}-x_{n-1} x_{n}+1\right) \\
\geq\left(\frac{3}{4}\right)^{n-1} \cdot\left(\left(\prod_{i=1}^{n} x_{i}\right)^{2}-\prod_{i=1}^{n} x_{i}+1\right)
\end{array}
$$
(2) If $x_{1}, x_{2}, \cdots, x_{n-1}$ are all not greater than 0. Then by the induction hypothesis, we have
$$
\begin{array}{l}
\prod_{i=1}^{n}\left(x_{i}^{2}-x_{i}+1\right) \geq \prod_{i=1}^{n-2}\left(x_{i}^{2}-x_{i}+1\right) \cdot \frac{3}{4}\left(\left(x_{n-1} x_{n}\right)^{2}-x_{n-1} x_{n}+1\right) \\
\geq\left(\frac{3}{4}\right)^{n-1} \cdot\left(\left(\prod_{i=1}^{n} x_{i}\right)^{2}-\prod_{i=1}^{n} x_{i}+1\right)
\end{array}
$$

By the above induction, the proposition is true.
In conclusion, the required $\lambda(n)=\left(\frac{3}{4}\right)^{n-1}$.
Note: When $n=3$, it is the 2006 USA Team Selection Test problem.