<p>Let \ \(a\) \ and \ \(b\) \ be two positive real numbers. Define:</p><p>\[A_1 = \frac{2a+b}{3},\quad A_2 = \frac{a+2b}{3}\]</p><p>\[G_1 = a^{2/3}b^{1/3},\quad G_2 = a^{1/3}b^{2/3}\]</p><p>\[H_1 = \frac{3ab}{a+2b},\quad H_2 = \frac{3ab}{2a+b}\]</p><p>Which of the following are correct?</p>
Step-by-Step Solution
Key Concept: Recognize that A₁, G₁, H₁ form an AM-GM-HM triple for weighted means, and similarly A₂, G₂, H₂ form another triple. The key is verifying the inequality chain AM ≥ GM ≥ HM holds for both groups, where equality occurs only when a = b.
<p><strong>Step 1: Verify the relationship for Group 1 (A₁, G₁, H₁)</strong></p><p>For positive reals a, b with weights 2/3 and 1/3:</p><p>• A₁ = (2a + b)/3 is the weighted arithmetic mean</p><p>• G₁ = a^(2/3)·b^(1/3) is the weighted geometric mean</p><p>• H₁ = 3ab/(2a+b) is the weighted harmonic mean</p><p><strong>Step 2: Apply weighted AM-GM inequality</strong></p><p>By AM-GM: (2a + b)/3 ≥ a^(2/3)·b^(1/3)</p><p>This gives A₁ ≥ G₁ ✓</p><p><strong>Step 3: Apply GM-HM inequality</strong></p><p>For weighted means: a^(2/3)·b^(1/3) ≥ 3ab/(2a+b)</p><p>This gives G₁ ≥ H₁ ✓</p><p><strong>Step 4: Verify for Group 2 (A₂, G₂, H₂)</strong></p><p>Similarly, with weights 1/3 and 2/3:</p><p>A₂ = (a + 2b)/3 ≥ a^(1/3)·b^(2/3) = G₂ ≥ 3ab/(a+2b) = H₂ ✓</p><p><strong>Step 5: Check equality conditions</strong></p><p>All inequalities become equalities if and only if a = b</p><p>∴ Answer: D</p>
Correct Answer: D