Basic Mathematics & Logarithm
Inequalities involving means
Grade 11
Question:
<p>If \(a, b, c \in R^+\), then \(\dfrac{bc}{b+c} + \dfrac{ac}{a+c} + \dfrac{ab}{a+b}\) is always</p>
<p>(1) \(\leq \dfrac{1}{2}(a+b+c)\)</p>
<p>(2) \(\geq \dfrac{1}{3}\sqrt{abc}\)</p>
<p>(3) \(\leq \dfrac{1}{3}(a+b+c)\)</p>
<p>(4) \(\geq \dfrac{1}{2}\sqrt{abc}\)</p>
Step-by-Step Solution
Key Concept: Recognize that each term has the form of harmonic mean relationship: xy/(x+y) = 1/(1/x + 1/y). Use the reciprocal substitution to convert to a sum of reciprocals, then apply AM-GM or direct inequality.
<p><strong>Step 1:</strong> Rewrite each term using reciprocals. Note that xy/(x+y) = 1/(1/x + 1/y).</p><p><strong>Step 2:</strong> Let S = bc/(b+c) + ac/(a+c) + ab/(a+b). Taking reciprocals: 1/S_term relates to harmonic means.</p><p><strong>Step 3:</strong> Use the substitution: (b+c)/bc = 1/b + 1/c, (a+c)/ac = 1/a + 1/c, (a+b)/ab = 1/a + 1/b.</p><p><strong>Step 4:</strong> By AM-GM inequality: 1/(1/a + 1/b) ≤ (a+b)/2, so ab/(a+b) ≤ (a+b)/4. Similarly for other terms.</p><p><strong>Step 5:</strong> Adding the inequalities: S ≤ (a+b)/4 + (b+c)/4 + (a+c)/4 = (a+b+c)/2.</p><p><strong>Step 6:</strong> Equality holds when a = b = c, giving S = a/2 at minimum. The expression is <strong>always ≤ (a+b+c)/2</strong> and <strong>≥ (a+b+c)/4</strong> by harmonic-arithmetic mean inequality.</p><p>∴ Answer: A</p>
Correct Answer: A