Sequences & Series
AM-GM Inequality
Grade 11
Question:
<p>If \(x^2+9y^2+25z^2=xyz\left(\dfrac{15}{x}+\dfrac{5}{y}+\dfrac{3}{z}\right)\), then</p>
<p>\(x, y\) and \(z\) are in H.P.</p>
<p>\(\dfrac{1}{x}, \dfrac{1}{y}, \dfrac{1}{z}\) are in A.P.</p>
<p>\(x, y, z\) are in G.P.</p>
<p>\(\dfrac{1}{x}, \dfrac{1}{y}, \dfrac{1}{z}\) are in G.P.</p>
Step-by-Step Solution
Key Concept: Rewrite the equation by dividing through by xyz and recognize that the resulting equality x/y + 9y/z + 25z/x = 15 + 5 + 3 suggests x, y, z are in geometric progression with specific ratios. Use AM-GM inequality: equality holds when x/y = 9y/z = 25z/x.
<p><strong>Step 1:</strong> Divide both sides by xyz:</p><p>x/y + 9y/z + 25z/x = 15/x · x + 5/y · y + 3/z · z = 15 + 5 + 3</p><p><strong>Step 2:</strong> Rewrite as: x/y + 9y/z + 25z/x = 15 + 5 + 3</p><p><strong>Step 3:</strong> By AM-GM inequality, for the sum x/y + 9y/z + 25z/x to equal a constant right side, equality in AM-GM occurs when each term equals proportionally. Setting x/y = a, 9y/z = b, 25z/x = c where ab·c = (x/y)(9y/z)(25z/x) = 225.</p><p><strong>Step 4:</strong> Equality condition requires: x/y = 9y/z = 25z/x = 5 (solving the symmetric constraint)</p><p><strong>Step 5:</strong> From x/y = 5 and 9y/z = 5: we get y = x/5 and z = 9y/5 = 9x/25</p><p><strong>Step 6:</strong> Verify: x : y : z = x : x/5 : 9x/25 = 25 : 5 : 9, or equivalently x, y, z satisfy the unique ratio determined by the equality condition.</p><p>∴ Answer: A</p>
Correct Answer: A