<p>If \(a, b, c\) are the sides of a triangle, then the minimum value of \(\dfrac{a}{b+c-a} + \dfrac{b}{c+a-b} + \dfrac{c}{a+b-c}\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the substitution x = b+c-a, y = c+a-b, z = a+b-c (which are all positive by triangle inequality). Then a, b, c can be expressed in terms of x, y, z, and the expression becomes a function of positive variables where x+y+z is constant.
<p><strong>Step 1:</strong> Let x = b+c-a, y = c+a-b, z = a+b-c where x, y, z > 0 (triangle inequality).</p><p><strong>Step 2:</strong> Express the variables: a = (y+z)/2, b = (x+z)/2, c = (x+y)/2.</p><p><strong>Step 3:</strong> The expression becomes: <br/>S = (y+z)/(2x) + (x+z)/(2y) + (x+y)/(2z)</p><p><strong>Step 4:</strong> Rewrite as: S = (1/2)[(y/x + x/y) + (z/x + x/z) + (z/y + y/z)]</p><p><strong>Step 5:</strong> By AM-GM inequality, for positive numbers: t/s + s/t ≥ 2 for each pair.</p><p><strong>Step 6:</strong> Therefore: S ≥ (1/2)[2 + 2 + 2] = 3</p><p><strong>Step 7:</strong> Equality holds when x = y = z, which means a = b = c (equilateral triangle).</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: A