Let $a_1=1$ and $a_{n+1}=2/(2+a_n)$ for all $n\geqslant 1$. Similarly, $b_1=1$ and $b_{n+1}=3/(3+b_n)$ for all $n\geqslant 1$. Which is greater between $a_{2022}$ and $b_{2022}$?

[i]Proposed by P. Kozhevnikov[/i]
1. **Define the sequences and their recurrence relations:**
   - For the sequence \( \{a_n\} \):
     \[
     a_1 = 1, \quad a_{n+1} = \frac{2}{2 + a_n} \quad \text{for all } n \geq 1
     \]
   - For the sequence \( \{b_n\} \):
     \[
     b_1 = 1, \quad b_{n+1} = \frac{3}{3 + b_n} \quad \text{for all } n \geq 1
     \]

2. **Generalize the sequences:**
   - Consider a generalized sequence \( \{u_n\} \) defined by:
     \[
     u_1 = 1, \quad u_{n+1} = \frac{u}{u + u_n} \quad \text{with } u > 1
     \]
   - The sequences \( \{a_n\} \) and \( \{b_n\} \) are special cases of \( \{u_n\} \) with \( u = 2 \) and \( u = 3 \) respectively.

3. **Find the fixed points of the recurrence relations:**
   - For \( \{a_n\} \):
     \[
     x = \frac{2}{2 + x} \implies x(2 + x) = 2 \implies x^2 + 2x - 2 = 0
     \]
     Solving this quadratic equation:
     \[
     x = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}
     \]
     Since \( x \) must be positive, we have:
     \[
     v_2 = \sqrt{3} - 1
     \]

   - For \( \{b_n\} \):
     \[
     x = \frac{3}{3 + x} \implies x(3 + x) = 3 \implies x^2 + 3x - 3 = 0
     \]
     Solving this quadratic equation:
     \[
     x = \frac{-3 \pm \sqrt{9 + 12}}{2} = \frac{-3 \pm \sqrt{21}}{2}
     \]
     Since \( x \) must be positive, we have:
     \[
     v_3 = \frac{-3 + \sqrt{21}}{2}
     \]

4. **Analyze the behavior of the sequences:**
   - Both sequences \( \{a_n\} \) and \( \{b_n\} \) converge to their respective fixed points \( v_2 \) and \( v_3 \).
   - The sequence \( \{a_n\} \) converges to \( v_2 = \sqrt{3} - 1 \).
   - The sequence \( \{b_n\} \) converges to \( v_3 = \frac{-3 + \sqrt{21}}{2} \).

5. **Compare the fixed points:**
   - Calculate \( v_2 \):
     \[
     v_2 = \sqrt{3} - 1 \approx 1.732 - 1 = 0.732
     \]
   - Calculate \( v_3 \):
     \[
     v_3 = \frac{-3 + \sqrt{21}}{2} \approx \frac{-3 + 4.583}{2} = \frac{1.583}{2} \approx 0.7915
     \]

6. **Conclusion:**
   - Since \( v_2 \approx 0.732 \) and \( v_3 \approx 0.7915 \), we have:
     \[
     v_2 < v_3
     \]
   - Therefore, \( a_{2022} \) converges to a value less than \( b_{2022} \).

The final answer is \( \boxed{ a_{2022} < b_{2022} } \)