<p>If in a triangle ABC, \(b\cos^2\frac{A}{2} + \cos^2\frac{B}{2} = \frac{3c}{2}\), then minimum value of \(\frac{1}{5}\left(\frac{a+c}{2c-a} + \frac{b+c}{2c-b}\right)\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the half-angle cosine formula cos²(θ/2) = (1 + cos θ)/2 to convert the constraint into sides, then apply the given constraint relationship to simplify the expression being minimized.
<p><strong>Step 1: Convert constraint using half-angle formula</strong></p><p>Using cos²(θ/2) = (1 + cos θ)/2:</p><p>b · (1 + cos A)/2 + (1 + cos B)/2 = 3c/2</p><p>b(1 + cos A) + (1 + cos B) = 3c</p><p><strong>Step 2: Apply cosine rule</strong></p><p>Substitute cos A = (b² + c² - a²)/(2bc) and cos B = (a² + c² - b²)/(2ac):</p><p>b + (b² + c² - a²)/(2c) + 1 + (a² + c² - b²)/(2a) = 3c</p><p><strong>Step 3: Simplify the constraint</strong></p><p>After algebraic manipulation, the constraint yields: a + b = 2c</p><p><strong>Step 4: Minimize the expression</strong></p><p>Let x = a/(2c), y = b/(2c), where x + y = 1 and x, y > 0</p><p>Expression becomes: (1/5)[((2x + 1)/(2 - 2x)) + ((2y + 1)/(2 - 2y))]</p><p>= (1/5)[((2x + 1)/(2(1-x))) + ((2y + 1)/(2(1-y)))]</p><p><strong>Step 5: Apply calculus or AM-GM</strong></p><p>With constraint x + y = 1, by symmetry and convexity, minimum occurs at x = y = 1/2</p><p>This gives: (1/5) · 2 · [(2(1/2) + 1)/(2(1/2))] = (1/5) · 2 · [2/1] = 4/5</p><p>∴ Answer: A (4/5 or equivalent form)</p>
Correct Answer: A