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📅 May 5, 2026 👤 scraface

May 05, 2026

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📌 5Question: Let \( a, b, c \) be positive real numbers such that \( a + b + c = 1 \). Find the minimum value of \( \frac{a}{1 - a} + \frac{b}{1 - b} + \frac{c}{1 - c} \).

📌 Solution: We are given \( a + b + c = 1 \) with \( a, b, c > 0 \), so \( 1 - a = b + c \), \( 1 - b = a + c \), and \( 1 - c = a + b \). The expression becomes:

📌 \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b}

📌 This is a well-known inequality in symmetric form. By the **Nesbitt’s Inequality**, for positive real numbers \( a, b, c \),

📌 \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \geq \frac{3}{2}

📌 with equality if and only if \( a = b = c \). Since \( a + b + c = 1 \), equality occurs when \( a = b = c = \frac{1}{3} \). Substituting, each term becomes:

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