6.25 $x, y>0, x, y \neq 1, \log _{x} y+\log _{y} x$ The range (in interval form) is
(A) $(-\infty,+\infty)$.
(B) $(-\infty,-2] \cup[2,+\infty)$.
(C) $(-\infty,-2) \cup[2,+\infty)$.
(D) $(-\infty,-2) \cup[2,+\infty)$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
［Solution］Consider the following two cases:
When $x>1, y>1$ or $0<x<1, 0<y<1$, $\log _{x} y>0$, $\log _{y} x>0$,
thus $\log _{x} y+\log _{y} x \geqslant 2 \sqrt{\log _{x} y \cdot \log _{y} x}=2$;
When $0<x<1, y>1$ or $x>1,0<y<1$,
$$
\log _{x} y<0, \quad \log _{y} x<0,
$$

at this time $-\left(\log _{x} y+\log _{y} x\right) \geqslant 2$, i.e., $\log _{x} y+\log _{y} x \leqslant-2$.
In summary, $\log _{x} y+\log _{y} x \in(-\infty,-2] \cup[2,+\infty)$.
Therefore, the answer is $(B)$.