<p>For any \(x, y \in R\), \(xy > 0\). Then the minimum value of \(\dfrac{2x}{y^3} + \dfrac{x^3 y}{3} + \dfrac{4y^2}{9x^4}\) is ______.</p>
Step-by-Step Solution
Key Concept: Recognize this as an optimization problem where AM-GM inequality applies after identifying equal weights. The expression has three terms that can be made equal at the minimum point by choosing appropriate substitution or by applying weighted AM-GM strategically.
<p><strong>Step 1:</strong> Given <strong>xy > 0</strong>, we need to minimize <strong>f(x,y) = 2x/y³ + x³y/3 + 4y²/(9x⁴)</strong>.</p><p><strong>Step 2:</strong> Apply AM-GM inequality to the three terms. For minimum, the terms must be equal at critical point:</p><p><strong>2x/y³ = x³y/3 = 4y²/(9x⁴)</strong></p><p><strong>Step 3:</strong> From first two terms: <strong>2x/y³ = x³y/3</strong></p><p>⟹ <strong>6x = x³y⁴</strong> ⟹ <strong>y⁴ = 6/x²</strong></p><p><strong>Step 4:</strong> From second and third terms: <strong>x³y/3 = 4y²/(9x⁴)</strong></p><p>⟹ <strong>9x⁷y = 12y²</strong> ⟹ <strong>9x⁷ = 12y</strong> ⟹ <strong>y = 3x⁷/4</strong></p><p><strong>Step 5:</strong> Solve simultaneously: <strong>(3x⁷/4)⁴ = 6/x²</strong></p><p>⟹ <strong>81x²⁸/256 = 6/x²</strong> ⟹ <strong>81x³⁰ = 1536</strong> ⟹ <strong>x³⁰ = 1536/81</strong></p><p><strong>Step 6:</strong> By AM-GM: <strong>f(x,y) = 3 × (2x/y³) = 3 × (4y²/(9x⁴)) = 3 × (x³y/3)</strong></p><p>Computing with equality condition gives each term = <strong>2/3</strong>, so minimum = <strong>3 × (2/3) = 2</strong></p><p>∴ <strong>Answer: 2</strong></p>
Correct Answer: 2