In a triangle the maximum value of `cos A +cos B+cos C` is.....

Answer

To find the maximum value of \( \cos A + \cos B + \cos C \) in a triangle, we can use some properties of triangles and trigonometric identities.

### Step-by-Step Solution:

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

2. **Using Trigonometric Identity**:
We can express \( \cos A + \cos B + \cos C \) using the identity:
\[ \cos A + \cos B + \cos C = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]
This identity relates the sum of the cosines of the angles to the product of the sines of half the angles.

3. **Applying AM-GM Inequality**:
To find the maximum value of \( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2}}{3} \geq \sqrt[3]{\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} \]

4. **Finding Maximum of Sine Terms**:
Since \( A + B + C = \pi \), we can use the fact that the maximum value of \( \sin x \) occurs at \( x = \frac{\pi}{2} \). Thus, the maximum value of \( \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \) occurs when \( A = B = C = \frac{\pi}{3} \).

5. **Calculating the Maximum Value**:
When \( A = B = C = \frac{\pi}{3} \):
\[ \sin \frac{A}{2} = \sin \frac{\pi/3}{2} = \sin \frac{\pi}{6} = \frac{1}{2} \]
Therefore,
\[ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

6. **Substituting Back to Find Cosine Sum**:
Now substituting back into the cosine identity:
\[ \cos A + \cos B + \cos C = 1 + 4 \cdot \frac{1}{8} = 1 + \frac{1}{2} = \frac{3}{2} \]

### Conclusion:
Thus, the maximum value of \( \cos A + \cos B + \cos C \) in a triangle is \( \frac{3}{2} \).

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