In a `DeltaABC,` `sin A sinB sinC` is

Options

A. `ge(3sqrt3)/(8)`
B. `le(3sqrt3)/(8)`
C. `le(sqrt3)/(8)`
D. `le(3)/(8)`

Explanation

To solve the problem of finding the value of \( \sin A \sin B \sin C \) in triangle \( ABC \), we can follow these steps:

### Step 1: Use the Identity for Sine Squared
We start by using the identity for sine squared:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 1 - \frac{1}{2}(\cos 2A + \cos 2B) + \sin^2 C \]
We can express \( \sin^2 A \) and \( \sin^2 B \) using the formula:
\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \]
Thus, we rewrite:
\[ \sin^2 A = \frac{1 - \cos 2A}{2}, \quad \sin^2 B = \frac{1 - \cos 2B}{2}, \quad \sin^2 C = \sin^2 C \]

### Step 2: Combine the Terms
Combining these terms gives:
\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{1 - \cos 2A}{2} + \frac{1 - \cos 2B}{2} + \sin^2 C \]
Factoring out \( \frac{1}{2} \):
\[ = \frac{1}{2}(2 - \cos 2A - \cos 2B) + \sin^2 C \]

### Step 3: Use the Cosine Addition Formula
Using the cosine addition formula:
\[ \cos 2A + \cos 2B = 2 \cos(A + B) \cos(A - B) \]
This allows us to rewrite:
\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{1}{2}(2 - 2 \cos(A + B) \cos(A - B)) + \sin^2 C \]

### Step 4: Simplify the Expression
This simplifies to:
\[ = 1 - \cos(A + B) \cos(A - B) + \sin^2 C \]
Since \( A + B + C = 180^\circ \), we have \( C = 180^\circ - (A + B) \).

### Step 5: Apply the Arithmetic Mean-Geometric Mean Inequality
Using the AM-GM inequality:
\[ \frac{\sin^2 A + \sin^2 B + \sin^2 C}{3} \geq \sqrt[3]{\sin^2 A \sin^2 B \sin^2 C} \]
Letting \( x = \sin A \sin B \sin C \), we have:
\[ \frac{1}{3} \geq \sqrt[3]{x^2} \]

### Step 6: Solve for \( x \)
Cubing both sides gives:
\[ \frac{1}{27} \geq x^2 \implies x \leq \frac{1}{3\sqrt{3}} \implies \sin A \sin B \sin C \leq \frac{3\sqrt{3}}{8} \]

### Final Result
Thus, we conclude:
\[ \sin A \sin B \sin C \leq \frac{3\sqrt{3}}{8} \]