1. Given that $x, y$ are two unequal positive numbers, then $A=$ $\sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2}, B=\frac{x+y}{2}-\sqrt{x y}, C=\sqrt{x y}-$ $\frac{2}{\frac{1}{x}+\frac{1}{y}}$
(A) $A>B>C$
(B) $A>C>B$
(C) $B>A>C$
(D) $B>C>A$
$-1 .(\mathrm{C})$
$$
\begin{aligned}
A-B & =\sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y}-(x+y) \\
& \leq \sqrt{2\left(\frac{x^{2}+y^{2}}{2}+x y\right)}-(x+y)=0 .
\end{aligned}
$$
$\because x, y$ are not equal, hence the equality cannot be achieved, $\therefore B>A$.
Also, $A-C$
$$
\begin{array}{l}
=\left(\sqrt{\frac{x^{2}+y^{2}}{2}}-\sqrt{x y}\right)-\left(\frac{x+y}{2}-\frac{2}{\frac{1}{x}+\frac{1}{y}}\right) \\
=\frac{\frac{1}{2}(x-y)^{2}}{\sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y}}-\frac{\frac{1}{2}(x-y)^{2}}{x+y}>0,
\end{array}
$$

Therefore, $B>A>C$.