<p>If <span>\( x_1, x_2, x_3, x_4, x_5 \)</span> are positive reals, find the minimum value of <span>\( \dfrac{x_1 + 2x_2 + 3x_3 + 4x_4 + 5x_5}{15} \)</span> given that it is <span>\( \geq (x_1 \cdot x_2^2 \cdot x_3^3 \cdot x_4^4 \cdot x_5^5)^{1/15} \)</span>. Find the value of <span>\(x_1 + x_2 + x_3 + x_4 + x_5\)</span> at equality.</p>
Step-by-Step Solution
Key Concept: Apply AM-GM inequality in weighted form: the weighted arithmetic mean equals the weighted geometric mean only when all terms are equal. Since weights are 1, 2, 3, 4, 5 (summing to 15), equality holds when x₁ = x₂ = x₃ = x₄ = x₅ = k for some constant k.
<p><strong>Step 1:</strong> Recognize the weighted AM-GM inequality: For positive reals and weights w₁, w₂, ..., wₙ summing to W:</p><p>$$\frac{w_1x_1 + w_2x_2 + ... + w_nx_n}{W} \geq (x_1^{w_1} \cdot x_2^{w_2} \cdot ... \cdot x_n^{w_n})^{1/W}$$</p><p><strong>Step 2:</strong> Here, weights are 1, 2, 3, 4, 5 with W = 15. Equality in weighted AM-GM holds when:</p><p>$$\frac{x_1}{1} = \frac{x_2}{2} = \frac{x_3}{3} = \frac{x_4}{4} = \frac{x_5}{5} = k$$</p><p>This gives: x₁ = k, x₂ = 2k, x₃ = 3k, x₄ = 4k, x₅ = 5k</p><p><strong>Step 3:</strong> Calculate the minimum value by substituting into the LHS:</p><p>$$\frac{k + 2(2k) + 3(3k) + 4(4k) + 5(5k)}{15} = \frac{k + 4k + 9k + 16k + 25k}{15} = \frac{55k}{15}$$</p><p><strong>Step 4:</strong> From the constraint inequality at equality, find k. The RHS becomes:</p><p>$$(k^1 \cdot (2k)^2 \cdot (3k)^3 \cdot (4k)^4 \cdot (5k)^5)^{1/15} = k(2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5)^{1/15}$$</p><p><strong>Step 5:</strong> For the minimum, the equality condition gives us the constraint. Since the answer asks for x₁ + x₂ + x₃ + x₄ + x₅:</p><p>$$x_1 + x_2 + x_3 + x_4 + x_5 = k + 2k + 3k + 4k + 5k = 15k$$</p><p>When k = 1/3 (from AM-GM equality constraint with the given form), this sum equals 5.</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: 5