1. Define the sets and polynomials:
Let $M = \{1, 2, \ldots, n\}$ be a set where each element is colored either red, blue, or yellow. Define the sets:
$$A = \{(x, y, z) \in M \times M \times M \mid x + y + z \equiv 0 \pmod{n}, \text{ and } x, y, z \text{ are of the same color} \}$$
$$B = \{(x, y, z) \in M \times M \times M \mid x + y + z \equiv 0 \pmod{n}, \text{ and } x, y, z \text{ are of pairwise distinct colors} \}$$

2. Introduce the polynomials:
Define the polynomials $R(x), B(x), Y(x)$ where:
$$R(x) = \sum_{r \in \text{Red}} x^r, \quad B(x) = \sum_{b \in \text{Blue}} x^b, \quad Y(x) = \sum_{y \in \text{Yellow}} x^y$$
Here, $r, b, y$ are elements of $M$ colored red, blue, and yellow respectively.

3. Use roots of unity:
Let $\omega$ be a primitive $n$-th root of unity. For $k \not\equiv 0 \pmod{n}$, we have:
$$R(\omega^k) + B(\omega^k) + Y(\omega^k) = \omega^k + \omega^{2k} + \cdots + \omega^{nk} = 0$$
This follows from the fact that the sum of all $n$-th roots of unity is zero.

4. Analyze the polynomials:
Each term in $R(x)^3$ represents a triplet from the set of red points. Specifically, the term $c_k x^k$ indicates there are $c_k$ red triplets with sum $k$. The same applies to $B(x)^3$ and $Y(x)^3$.

5. Consider the product polynomial:
The polynomial $R(x) \cdot B(x) \cdot Y(x)$ gives the number of ordered triplets $(x, y, z)$ with colors red, blue, and yellow respectively.

6. Apply Euler's identity:
Using Euler's identity, we have:
$$R(x)^3 + B(x)^3 + Y(x)^3 - 3R(x)B(x)Y(x) = \frac{1}{2} \left( R(x) + B(x) + Y(x) \right) \left( (R(x) - B(x))^2 + (B(x) - Y(x))^2 + (Y(x) - R(x))^2 \right)$$

7. Combine the results:
For $k \not\equiv 0 \pmod{n}$, combining the results, we get:
$$R(\omega^k)^3 + B(\omega^k)^3 + Y(\omega^k)^3 = 3R(\omega^k)B(\omega^k)Y(\omega^k)$$
When $k = n$, it becomes an inequality because $R(1), B(1), Y(1) \in \mathbb{N}$.

8. Use the root of unity filter:
By the root of unity filter, we have:
$$\frac{1}{n} \sum_{k=0}^{n-1} \left( R(\omega^k)^3 + B(\omega^k)^3 + Y(\omega^k)^3 - 3R(\omega^k)B(\omega^k)Y(\omega^k) \right) = |A| - 3|B'| \geq 0$$
Here, $B'$ is the set of ordered triplets with colors red, blue, and yellow.

9. **Relate $B$ and $B'$:**
Each triplet in $B'$ can be permuted in $3! = 6$ ways to get a triplet in $B$. Thus, $|B'| = \frac{1}{6} |B|$.

10. Conclude the inequality:
Substituting $|B'| = \frac{1}{6} |B|$ into the inequality, we get:
$$|A| - 3 \left( \frac{1}{6} |B| \right) \geq 0 \implies |A| \geq \frac{1}{2} |B|$$
Therefore, we have:
$$2|A| \geq |B|$$

The final answer is $\boxed{ 2|A| \geq |B| }$.