6. 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 $\sum_{j=1}^{n} \alpha_{j}^{2}<1, \sum_{j=1}^{n} \beta_{j}^{2}<1$. Denote
$$
A^{2}=1-\sum_{j=1}^{n} \alpha_{j}^{2}, B^{2}=1-\sum_{j=1}^{n} \beta_{j}^{2}, W=\frac{1}{2}\left(1-\sum_{j=1}^{n} \alpha_{j} \beta_{j}\right)^{2} .
$$

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.
6. Obviously, when $\lambda=0$, the equation in the problem has only real roots.

Now consider the case where $\lambda \neq 0$. In this case, the equation has no zero roots. Suppose the $n$ roots of the equation are all real, denoted as $\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n}$, where $\varepsilon_{j} \in \mathbf{R} \backslash\{0\}, j=1,2, \cdots, n$. Now observe the relationship between the roots and the coefficients:
$$
\begin{array}{l}
\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{n}=(-1)^{n} \lambda, \\
\varepsilon_{1} \varepsilon_{2} \cdots \varepsilon_{n}\left(\frac{1}{\varepsilon_{1}}+\frac{1}{\varepsilon_{2}}+\cdots+\frac{1}{\varepsilon_{n}}\right)=(-1)^{n-1} \lambda A B, \\
\varepsilon_{1} \varepsilon_{2} \cdots \varepsilon_{n}\left(\sum_{1 \leqslant j0 .
$$

However,
$$
\begin{aligned}
A^{2} B^{2} & =\left(1-\sum_{j=1}^{n} \alpha_{j}^{2}\right)\left(1-\sum_{j=1}^{n} \beta_{j}^{2}\right) \\
& \leqslant\left[\frac{\left(1-\sum_{j=1}^{n} \alpha_{j}^{2}\right)+\left(1-\sum_{j=1}^{n} \beta_{j}^{2}\right)}{2}\right]^{2}=\left[1-\frac{1}{2} \sum_{j=1}^{n}\left(\alpha_{j}^{2}+\beta_{j}^{2}\right)\right]^{2} \\
& \leqslant\left(1-\sum_{j=1}^{n} \alpha_{j} \beta_{j}\right)^{2}=2 W .
\end{aligned}
$$

The contradiction derived above indicates that any $\lambda \neq 0$ does not meet the requirement.
Therefore, for the equation in the problem to have only real roots, it is necessary and sufficient that $\lambda=0$.