4.2.5 ** Let $x, y$ be two distinct positive numbers,
$$
Q=\sqrt{\frac{x^{2}+y^{2}}{2}}, A=\frac{x+y}{2}, G=\sqrt{x y}, H=\frac{2 x y}{x+y} .
$$

Determine which of $Q-A, A-G, G-H$ is the largest and which is the smallest.
Parse the relationship among these three numbers: 
\[ G-H\sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2} \Leftrightarrow x+y>\sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y} \\
\Leftrightarrow(x+y)^{2}>\frac{x^{2}+y^{2}}{2}+x y+\sqrt{2 x y\left(x^{2}+y^{2}\right)} \\
\quad \Leftrightarrow \frac{1}{2}(x+y)^{2}>\sqrt{2 x y\left(x^{2}+y^{2}\right)} \Leftrightarrow(x+y)^{4}>8 x y\left(x^{2}+y^{2}\right) \\
\Leftrightarrow x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4} \geqslant 8 x^{3} y+8 x y^{3} \\
\Leftrightarrow(x-y)^{4}>0 .
\end{array}
\]

Since \( x, y \) are not equal, the last inequality above holds, so \( A-G > Q-A \). Next, we prove: \( Q-A > G-H \).
\[
\begin{aligned}
\sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2} & >\sqrt{x y}-\frac{2 x y}{x+y} \Leftrightarrow \sqrt{\frac{x^{2}+y^{2}}{2}}-\sqrt{x y}>\frac{x+y}{2}-\frac{2 x y}{x+y} \\
\Leftrightarrow & \frac{\frac{1}{2}\left(x^{2}+y^{2}\right)-x y}{\sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y}}>\frac{(x-y)^{2}}{2(x+y)} \Leftrightarrow x+y>\sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y .}
\end{aligned}
\]

The last inequality above has already been proven, so \( Q-A > G-H \).
In summary, \( A-G \) is the largest, and \( G-H \) is the smallest.