Let $ 0 < x <y$ be real numbers. Let
$ H\equal{}\frac{2xy}{x\plus{}y}$ , $ G\equal{}\sqrt{xy}$ , $ A\equal{}\frac{x\plus{}y}{2}$ , $ Q\equal{}\sqrt{\frac{x^2\plus{}y^2}{2}}$
be the harmonic, geometric, arithmetic and root mean square (quadratic mean) of $ x$ and $ y$. As generally known $ H<G<A<Q$. Arrange the intervals $ [H,G]$ , $ [G,A]$ and $ [A,Q]$ in ascending order by their length.
Given the harmonic mean \( H = \frac{2xy}{x+y} \), geometric mean \( G = \sqrt{xy} \), arithmetic mean \( A = \frac{x+y}{2} \), and root mean square (quadratic mean) \( Q = \sqrt{\frac{x^2 + y^2}{2}} \), we need to arrange the intervals \([H, G]\), \([G, A]\), and \([A, Q]\) in ascending order by their length.

1. **Calculate the lengths of the intervals:**
   - Length of \([H, G]\): \( G - H \)
   - Length of \([G, A]\): \( A - G \)
   - Length of \([A, Q]\): \( Q - A \)

2. **Express \( x \) and \( y \) in terms of \( k \):**
   Without loss of generality, assume \( x > y \). Let \( x = ky \) where \( k > 1 \).

3. **Calculate the means in terms of \( k \) and \( y \):**
   - \( H = \frac{2xy}{x+y} = \frac{2ky \cdot y}{ky + y} = \frac{2ky^2}{(k+1)y} = \frac{2ky}{k+1} \)
   - \( G = \sqrt{xy} = \sqrt{ky \cdot y} = \sqrt{ky^2} = y\sqrt{k} \)
   - \( A = \frac{x+y}{2} = \frac{ky + y}{2} = y \cdot \frac{k+1}{2} \)
   - \( Q = \sqrt{\frac{x^2 + y^2}{2}} = \sqrt{\frac{(ky)^2 + y^2}{2}} = \sqrt{\frac{k^2y^2 + y^2}{2}} = y \sqrt{\frac{k^2 + 1}{2}} \)

4. **Calculate the interval lengths:**
   - \( G - H = y\sqrt{k} - \frac{2ky}{k+1} = y \left( \sqrt{k} - \frac{2k}{k+1} \right) \)
   - \( A - G = y \cdot \frac{k+1}{2} - y\sqrt{k} = y \left( \frac{k+1}{2} - \sqrt{k} \right) \)
   - \( Q - A = y \sqrt{\frac{k^2 + 1}{2}} - y \cdot \frac{k+1}{2} = y \left( \sqrt{\frac{k^2 + 1}{2}} - \frac{k+1}{2} \right) \)

5. **Compare the interval lengths:**
   - To compare \( A - G \) and \( Q - A \):
     \[
     \frac{k+1}{2} - \sqrt{k} > \sqrt{\frac{k^2 + 1}{2}} - \frac{k+1}{2}
     \]
     This inequality holds because \( \sqrt{\frac{k^2 + 1}{2}} \leq \frac{k+1}{2} \) by the AM-QM inequality.

   - To compare \( Q - A \) and \( G - H \):
     \[
     \sqrt{\frac{k^2 + 1}{2}} - \frac{k+1}{2} > \sqrt{k} - \frac{2k}{k+1}
     \]
     This inequality holds because \( \sqrt{\frac{k^2 + 1}{2}} \geq \sqrt{k} \) and \( \frac{k+1}{2} \leq \frac{2k}{k+1} \).

6. **Conclusion:**
   \[
   A - G \geq Q - A \geq G - H
   \]

The final answer is \( \boxed{ [G, H] \leq [Q, A] \leq [A, G] } \)