Determine all the positive real numbers $x_1, x_2, x_3, \dots, x_{2021}$ such that 
$x_{i+1}=\frac{x_i^3+2}{3x_i^2}$
for every $i=1, 2, 3, \dots, 2020$ and $x_{2021}=x_1$
To determine all the positive real numbers \( x_1, x_2, x_3, \dots, x_{2021} \) such that \( x_{i+1} = \frac{x_i^3 + 2}{3x_i^2} \) for every \( i = 1, 2, 3, \dots, 2020 \) and \( x_{2021} = x_1 \), we start by analyzing the function \( f(x) = \frac{x^3 + 2}{3x^2} \).

1. **Define the function and its properties:**
   \[
   f(x) = \frac{x^3 + 2}{3x^2} = \frac{1}{3} \left( x + \frac{2}{x^2} \right)
   \]
   The derivative of \( f(x) \) is:
   \[
   f'(x) = \frac{d}{dx} \left( \frac{1}{3} \left( x + \frac{2}{x^2} \right) \right) = \frac{1}{3} \left( 1 - \frac{4}{x^3} \right)
   \]
   The function \( f(x) \) is strictly decreasing on \( (0, \sqrt[3]{4}] \) and strictly increasing on \( [\sqrt[3]{4}, +\infty) \). The minimum value of \( f(x) \) is:
   \[
   f(\sqrt[3]{4}) = \frac{1}{\sqrt[3]{2}}
   \]

2. **Fixed points and behavior:**
   \[
   f(x) = x \Longleftrightarrow x = 1
   \]
   \[
   f(x) < x \text{ for } x > 1 \quad \text{and} \quad f(x) > x \text{ for } x \in (0, 1)
   \]
   \[
   f(x) = 1 \Longleftrightarrow x = 1 \text{ or } x = 1 + \sqrt{3}
   \]
   \[
   f(x) < 1 \Longleftrightarrow x \in (1, 1 + \sqrt{3}) \quad \text{and} \quad f(x) > 1 \Longleftrightarrow x \in (0, 1) \cup (1 + \sqrt{3}, +\infty)
   \]

3. **Case analysis:**
   - **Case 1: \( x_1 = 1 \)**
     \[
     x_n = 1 \quad \forall n \in \{1, 2, \dots, 2021\}
     \]
     This satisfies \( x_{2021} = x_1 \).

   - **Case 2: \( x_1 = 1 + \sqrt{3} \)**
     \[
     x_2 = f(1 + \sqrt{3}) = 1
     \]
     \[
     x_n = 1 \quad \forall n \in \{2, 3, \dots, 2021\}
     \]
     This does not satisfy \( x_{2021} = x_1 \).

   - **Case 3: \( x_1 > 1 + \sqrt{3} \)**
     \[
     x_2 < x_1
     \]
     - **Subcase 3.a: \( x_1 > x_2 > \dots > x_{2020} > x_{2021} \ge 1 + \sqrt{3} \)**
       \[
       x_{2021} < x_1
       \]
     - **Subcase 3.b: \( \exists k \in \{2, 3, \dots, 2021\} \) such that \( x_1 > x_2 > \dots > x_{k-1} > 1 + \sqrt{3} \) and \( 1 < x_k \le 1 + \sqrt{3} \)**
       \[
       x_n \le 1 + \sqrt{3} \quad \forall n \in \{k, k+1, \dots, 2021\}
       \]
       \[
       x_{2021} < x_1
       \]

   - **Case 4: \( x_1 \in (1, 1 + \sqrt{3}) \)**
     \[
     x_2 \in \left( \frac{1}{\sqrt[3]{2}}, 1 \right)
     \]
     \[
     x_3 \in \left( 1, \frac{5 \sqrt[3]{4}}{6} \right) \subset (1, \sqrt[3]{4})
     \]
     \[
     x_4, x_6, \dots, x_{2018}, x_{2020} \in \left( \frac{1}{\sqrt[3]{2}}, 1 \right)
     \]
     \[
     x_5, x_7, \dots, x_{2019}, x_{2021} \in (1, \sqrt[3]{4})
     \]
     \[
     x_1 > x_3 > \dots > x_{2019} > x_{2021} \Longrightarrow x_{2021} < x_1
     \]

   - **Case 5: \( x_1 \in \left[ \frac{1}{\sqrt[3]{2}}, 1 \right) \)**
     \[
     x_2, x_4, \dots, x_{2018}, x_{2020} \in \left( 1, \frac{5 \sqrt[3]{4}}{6} \right]
     \]
     \[
     x_3, x_5, \dots, x_{2019}, x_{2021} \in \left( \frac{1}{\sqrt[3]{2}}, 1 \right)
     \]
     \[
     x_1 < x_3 < \dots < x_{2019} < x_{2021} \Longrightarrow x_{2021} > x_1
     \]

   - **Case 6: \( x_1 \in \left( 0, \frac{1}{\sqrt[3]{2}} \right) \)**
     \[
     x_2 > \frac{5 \sqrt[3]{4}}{6}
     \]
     - **Subcase 6.a: \( x_2 > x_3 > \dots > x_{2020} > x_{2021} > 1 + \sqrt{3} \)**
       \[
       x_{2021} > x_1
       \]
     - **Subcase 6.b: \( \exists k \in \{3, 4, \dots, 2021\} \) such that \( x_2 > x_3 > \dots > x_{k-1} > 1 + \sqrt{3} \) and \( 1 < x_k \le 1 + \sqrt{3} \)**
       \[
       x_n \ge \frac{1}{\sqrt[3]{2}} \quad \forall n \in \{k, k+1, \dots, 2021\}
       \]
       \[
       x_{2021} > x_1
       \]
     - **Subcase 6.c: \( 1 < x_2 \le 1 + \sqrt{3} \)**
       \[
       x_n \ge \frac{1}{\sqrt[3]{2}} \quad \forall n \in \{k, k+1, \dots, 2021\}
       \]
       \[
       x_{2021} > x_1
       \]

Hence, the condition \( x_{2021} = x_1 \) can occur only in the case \( x_1 = x_2 = \dots = x_{2021} = 1 \).

The final answer is \( \boxed{ x_1 = x_2 = \dots = x_{2021} = 1 } \)