9-133 Let $x, y$ be two distinct positive numbers, and let
$$
R=\sqrt{\frac{x^{2}+y^{2}}{2}}, A=\frac{x+y}{2}, G=\sqrt{x y}, H=\frac{2 x y}{x+y} .
$$

Determine the size relationship among $R-A, A-G, G-H$.
[Solution] Clearly $G<A$. Since $R=\sqrt{2 A^{2}-G^{2}}$, we have
$$
R-A=\sqrt{2 A^{2}-G^{2}}-A=\frac{(A+G)(A-G)}{\sqrt{2 A^{2}-G^{2}}+A}.
$$

From $G<A$, it immediately follows that
$$
A+G<A+\sqrt{2 A^{2}-G^{2}},
$$

thus we have $\quad R-A<A-G$.
From $G^{2}<2 A^{2}-G^{2}$, we know
$$
G \sqrt{2 A^{2}-G^{2}}=\sqrt{G^{2}\left(2 A^{2}-G^{2}\right)}<A^{2}.
$$

This leads to $A G+G \sqrt{2 A^{2}-G^{2}}<A G+A^{2}$, which means
$$
\frac{G}{A}<\frac{A+G}{\sqrt{2 A^{2}-G^{2}}+A}.
$$

Noting that $H=\frac{G^{2}}{A}$, we have $G-H=\frac{G}{A}(A-G)<R-A$.
In summary, we get $\quad G-H<R-A<A-G$.