Example 20. (i) In a AABC, find the maxima and minima of u = sin A sin B sin C

where A + B + C = T

(ii) In a plane triangle ABC, find the maximum value of cosA cosB cosC.

To tackle the problem you've presented, we need to break it down into two parts: first, finding the maxima and minima of the function \( u = \sin A \sin B \sin C \) under the constraint \( A + B + C = T \), and second, determining the maximum value of \( \cos A \cos B \cos C \) in a triangle. Let's dive into each part step by step.

Finding Maxima and Minima of \( u = \sin A \sin B \sin C \)

Given the constraint \( A + B + C = T \), we can express one variable in terms of the others. For instance, we can write \( C = T - A - B \). Substituting this into our function gives:

\( u = \sin A \sin B \sin(T - A - B) \)

To find the extrema of \( u \), we can use the method of Lagrange multipliers or directly differentiate with respect to \( A \) and \( B \). However, a more intuitive approach is to analyze the behavior of the sine function, which oscillates between -1 and 1.

Using Symmetry and Critical Points

Since \( \sin \) is symmetric and periodic, we can expect that the maximum occurs when \( A = B = C \). Setting \( A = B = C = \frac{T}{3} \) gives:

\( u = \sin\left(\frac{T}{3}\right) \sin\left(\frac{T}{3}\right) \sin\left(\frac{T}{3}\right) = \left(\sin\left(\frac{T}{3}\right)\right)^3 \)

To find the maximum value, we need to ensure that \( T \) is within a range where \( \sin\left(\frac{T}{3}\right) \) is maximized, typically when \( T \) is \( \pi \) or \( 2\pi \). The minimum value occurs when any angle approaches 0 or \( \pi \), leading \( u \) to approach 0.

Maximizing \( \cos A \cos B \cos C \)

For the second part, we want to maximize \( v = \cos A \cos B \cos C \) in a triangle where \( A + B + C = \pi \). Similar to the sine case, we can use the symmetry of the cosine function.

Applying the AM-GM Inequality

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know:

• For any non-negative \( x, y, z \): \( \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \)

Applying this to \( \cos A, \cos B, \cos C \), we have:

\( \frac{\cos A + \cos B + \cos C}{3} \geq \sqrt[3]{\cos A \cos B \cos C} \end{p}

To maximize \( \cos A \cos B \cos C \), we set \( A = B = C = \frac{\pi}{3} \). Thus:

\( v = \cos\left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right) \cos\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \end{p}

In summary, the maximum value of \( \cos A \cos B \cos C \) in a triangle is \( \frac{1}{8} \), occurring when each angle is \( \frac{\pi}{3} \). The minimum value, on the other hand, approaches 0 when any angle approaches \( \frac{\pi}{2} \) or \( 0 \).

Both parts of your question illustrate the beauty of trigonometric functions and their symmetrical properties in geometric contexts. If you have any further questions or need clarification on any steps, feel free to ask!