Consider the problem: Let $n \geqslant 4, \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}, \beta_{1}, \beta_{2}, \cdots, \beta_{n}$ be two sets of real numbers, satisfying $\sum_{i=1}^{n} \alpha_{i}^{2}<1, \sum_{i=1}^{n} \beta_{i}^{2}<1$, and let $A^{2}=1-\sum_{i=1}^{n} \alpha_{i}^{2}, B^{2}=1-\sum_{i=1}^{n} \beta_{i}^{2}, W=\frac{1}{2}\left(1-\sum_{i=1}^{n} \alpha_{i} \beta_{i}\right)^{2}$. Find all real numbers $\lambda$ such that the equation $x^{n}+\lambda$ $\left(x^{n-1}+x^{n-2}+\cdots+x^{3}+W x^{2}+A B x+1\right)=0$ has only real roots.
Prove that when $\lambda=0$, the equation has only the real root 0.
When $\lambda \neq 0$, $\Delta_{2}=(n-1)(\lambda A B)^{2}-2 n(\lambda W) \lambda=\lambda^{2}\left[(n-1) A^{2} B^{2}-2 n W\right)$.
From $1-\sqrt{2 W}=\sum_{i=1}^{n} \alpha_{1} \beta_{i} \leqslant \sum_{i=1}^{n} \frac{1}{2}\left(\alpha_{i}^{2}+\beta_{i}^{2}\right)=\frac{2-A^{2}-B^{2}}{2}$,
we get $\sqrt{2 W} \geqslant \frac{A^{2}+B^{2}}{2} \geqslant A B$. Therefore, $2 W \geqslant A^{2} B^{2}$. Hence,
$$
\Delta_{2} \leqslant \lambda^{2}\left[2(n-1) A^{2} B^{2}-2 n A^{2} B^{2}\right]=-2 A^{2} B^{2} \lambda^{2}<0 .
$$

By Lemma (2) of Example 10, the roots of the equation are not all real.
In summary, it can only be that $\lambda=0$.