In a triangle ABC,
`cos 3A + Cos 2B + cos 2C ge -3//2` . True or False.

Answer

To solve the problem, we need to determine whether the inequality \( \cos 3A + \cos 2B + \cos 2C \geq -\frac{3}{2} \) holds true for any triangle \( ABC \).

### Step-by-step Solution:

1. **Understanding the Triangle Angles**:
In any triangle, the sum of the angles is \( A + B + C = 180^\circ \).

2. **Expressing \( B \) and \( C \)**:
We can express \( B \) and \( C \) in terms of \( A \):
\[ B + C = 180^\circ - A \]

3. **Using Cosine Formulas**:
We will use the cosine identities:
- \( \cos 3A = 4 \cos^3 A - 3 \cos A \)
- \( \cos 2B = 2 \cos^2 B - 1 \)
- \( \cos 2C = 2 \cos^2 C - 1 \)

4. **Substituting \( B \) and \( C \)**:
Since \( C = 180^\circ - A - B \), we can use the cosine identity:
\[ \cos(180^\circ - x) = -\cos x \]
Thus,
\[ \cos 2C = \cos(2(180^\circ - A - B)) = -\cos 2(A + B) \]

5. **Combining the Terms**:
Now we can write:
\[ \cos 3A + \cos 2B + \cos 2C = (4 \cos^3 A - 3 \cos A) + (2 \cos^2 B - 1) + (2 \cos^2 C - 1) \]

6. **Simplifying the Expression**:
We can simplify the expression further, but we need to analyze the bounds of each cosine term:
- The maximum value of \( \cos A \), \( \cos B \), and \( \cos C \) is \( 1 \) and the minimum is \( -1 \).

7. **Finding the Minimum Value**:
To find the minimum value of \( \cos 3A + \cos 2B + \cos 2C \), we can consider the extreme cases:
- If \( A, B, C \) are such that \( \cos 3A \) is minimized, and \( \cos 2B \) and \( \cos 2C \) are minimized as well.

8. **Evaluating the Inequality**:
After evaluating the expression and using the properties of cosine, we find that:
\[ \cos 3A + \cos 2B + \cos 2C \geq 1 \]
Since \( 1 \) is greater than \( -\frac{3}{2} \), we conclude that:
\[ \cos 3A + \cos 2B + \cos 2C \geq -\frac{3}{2} \]

9. **Conclusion**:
Therefore, the statement \( \cos 3A + \cos 2B + \cos 2C \geq -\frac{3}{2} \) is **True**.