Example 2 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
$$
\sum_{i=1}^{n} \alpha_{i}^{2}<1, \sum_{i=1}^{n} \beta_{i}^{2}<1 .
$$

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
$$
x^{n}+\lambda\left(x^{n-1}+\cdots+x^{3}+W x^{2}+A B x+1\right)=0
$$

has only real roots.
When $\lambda=\hat{*}$, the equation has only the real root 0.
When $\lambda \neq \neq$ (),
$$
\begin{aligned}
\Delta_{i} & ==(n-1)(\lambda A B)^{2}-2 n(\lambda W) \lambda \\
& =\lambda^{2}\left[(n-1) A^{2} B^{2}-2 n W\right] .
\end{aligned}
$$

By $1-\sqrt{2 \bar{W}}=\sum_{i=1}^{n} \alpha_{i} \beta_{i} \leqslant \sum_{i=1}^{n} \frac{1}{2}\left(\alpha_{i}^{2}+\beta_{i}^{2}\right)$
$$
=\frac{2-A^{2}-B^{2}}{2} \text {, }
$$

we get $\sqrt{2 W} \geqslant \frac{A^{2}+B^{2}}{2} \geqslant A B$.
Therefore, $2 W \geqslant A^{2} B^{2}$.
Thus, $\Delta_{2} \leqslant \lambda^{2}\left[(n-1) A^{2} B^{2}-n A^{2} B^{2}\right]$ $=-A^{2} B^{2} \lambda^{2}<0$. By Theorem (2), we know that the roots of the equation are not all real.
In summary, the only possibility is $\lambda=0$.