1. Given $a=\sqrt{2}-1, b=2 \sqrt{2}-\sqrt{6}, c=\sqrt{6}-2$. Then, the size relationship of $a, b, c$ is ( ).
(A) $a<b<c$
(B) $b<a<c$
(C) $c<b<a$
(D) $c<a<b$
- 1.(B).

Since $3>2 \sqrt{2}$, then $a-b=\sqrt{6}-1-\sqrt{2} \geqslant \sqrt{3+2 \sqrt{2}}-1$ $-\sqrt{2}=0$. Similarly, $c-a>0$. Therefore, $b<a<c$.