Let $x_0, x_1, x_2 \dots$ be a sequence of positive real numbers such that for all $n \geq 0$, $$x_{n+1} = \dfrac{(n^2+1)x_n^2}{x_n^3+n^2}$$ For which values of $x_0$ is this sequence bounded?
To determine for which values of \( x_0 \) the sequence \( \{x_n\} \) is bounded, we analyze the given recurrence relation:
\[ x_{n+1} = \frac{(n^2+1)x_n^2}{x_n^3 + n^2} \]

1. **Claim 1**: If \( x_n \leq 1 \) for \( n \geq 1 \), then \( x_{n+1} \leq 1 \).

   **Proof**: We need to show that \( x_{n+1} \leq 1 \) when \( x_n \leq 1 \). Consider the expression:
   \[ x_{n+1} \leq 1 \]
   This is equivalent to:
   \[ \frac{(n^2+1)x_n^2}{x_n^3 + n^2} \leq 1 \]
   Simplifying, we get:
   \[ (n^2+1)x_n^2 \leq x_n^3 + n^2 \]
   Rearranging terms, we obtain:
   \[ x_n^3 - (n^2+1)x_n^2 + n^2 \geq 0 \]
   Factoring out \( x_n^2 \), we have:
   \[ x_n^2(x_n - (n^2+1)) + n^2 \geq 0 \]
   Since \( x_n \leq 1 \), the term \( x_n - (n^2+1) \) is negative, and thus the inequality holds. Therefore, \( x_{n+1} \leq 1 \) if \( x_n \leq 1 \). \(\blacksquare\)

2. **Claim 2**: If \( x_n \leq 1 \) for some \( n \geq 1 \), then either \( x_1 \leq 1 \) or \( x_2 \leq 1 \).

   **Proof**: Assume for the sake of contradiction that the smallest positive integer \( n \) such that \( x_n > 1 \) and \( x_{n+1} \leq 1 \) is at least 2. Using \( x_n > 1 \) in the expression for \( x_{n+1} \leq 1 \), we get:
   \[ x_n \geq \frac{n^2 + n\sqrt{n^2 + 4}}{2} > n^2 \]
   Now, using \( x_{n-1} > 1 \) in the expression for \( x_n \), we have:
   \[ x_n = \frac{(n-1)^2 + 1}{x_{n-1} + \frac{(n-1)^2}{x_{n-1}^2}} < (n-1)^2 + 1 \]
   This leads to a contradiction. Therefore, if \( x_n \leq 1 \) for some \( n \geq 1 \), then either \( x_1 \leq 1 \) or \( x_2 \leq 1 \). \(\blacksquare\)

3. Solving for \( x_1 \leq 1 \) and \( x_2 \leq 1 \) gives:
   \[ \boxed{x_0 \in \left(0, \frac{-1 + \sqrt{5}}{2}\right] \cup [1, +\infty)} \]
   Clearly, these values work.

4. **Claim 3**: If \( x_n > 1 \) for all sufficiently large \( n \) and \( \{x_n\} \) is bounded, then \( \{x_n\} \) is eventually strictly increasing.

   **Proof**: Write:
   \[ x_{n+1} - x_n = \frac{x_n(x_n - 1)(n^2 - x_n - x_n^2)}{x_n^3 + n^2} \]
   Since \( x_n > 1 \) eventually, and \( \{x_n\} \) bounded implies \( x_n^2 + x_n < n \) for all sufficiently large \( n \), \( \{x_n\} \) is eventually strictly increasing. \(\blacksquare\)

5. Now assume we have a sequence \( \{x_n\} \) which is bounded but is eventually \( > 1 \) always. Then it is eventually strictly increasing, plus bounded so it has a limit \( L > 1 \). But, rewrite:
   \[ x_{n+1} = \frac{\left(1 + \frac{1}{n^2}\right)x_n^2}{1 + \frac{x_n^3}{n^2}} \]
   Taking \( n \rightarrow \infty \) in the above, we get:
   \[ L = L^2 \]
   which is impossible since \( L > 1 \).

Therefore, every such bounded sequence must stay below 1 eventually, and Claim 2 gives us the required values of \( x_0 \) that work. \(\blacksquare\)