Example 4 Let $x, y$ be two distinct real numbers, $R=\sqrt{\frac{x^{2}+y^{2}}{2}}, A=\frac{x+y}{2}, G$ $=\sqrt{x y}, H=\frac{2 x y}{x+y}$, determine which of $R-A, A-G, G-H$ is the largest and which is the smallest. (30th IMO Canadian Training Problem)
Solve $A-G$ is maximum, $G-H$ is minimum. Because
$$\begin{array}{l} 
\frac{x+y}{2}-\sqrt{x y} \geqslant \sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2} \\
\Leftrightarrow x+y \geqslant \sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y} \\
\Leftrightarrow(x+y)^{2} \geqslant \frac{x^{2}+y^{2}}{2}+x y+\sqrt{2 x y\left(x^{2}+y^{2}\right)} \\
\Leftrightarrow \frac{(x+y)^{2}}{2} \geqslant \sqrt{2 x y\left(x^{2}+y^{2}\right)} \\
\Leftrightarrow(x+y)^{4} \geqslant 8 x y\left(x^{2}+y^{2}\right) \Leftrightarrow(x-y)^{4} \geqslant 0 .
\end{array}$$

The last inequality is obviously true, so $A-G \geqslant R-A$.
Also, because $\sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2} \geqslant \sqrt{x y}-\frac{2 x y}{x+y}$
$$\Leftrightarrow \sqrt{\frac{x^{2}+y^{2}}{2}}-\sqrt{x y} \geqslant \frac{x+y}{2}-\frac{2 x y}{x+y}=\frac{(x-y)^{2}}{2(x+y)} \geqslant 0,$$

so $R-A \geqslant G-H$.