<p>If <i>x</i>, <i>y</i>, <i>z</i> are positive numbers in A.P., then which of the following holds?</p>
<p>(1) \(y^2 \ge xz\)</p>
<p>(2) \(xy + yz \ge 2xz\)</p>
<p>(3) \(\dfrac{x+y}{2y-x} + \dfrac{y+z}{2y-z} \ge 4\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: When three positive numbers are in A.P., we have y - x = z - y, which means y = (x+z)/2. This relationship, combined with positivity constraints, allows us to test inequalities involving these terms using AM-GM or algebraic manipulation.
<p><strong>Step 1:</strong> Given x, y, z are in A.P. with x, y, z > 0, we have:</p><p>2y = x + z</p><p><strong>Step 2:</strong> For positive numbers in A.P., test the reciprocals. If x, y, z are in A.P., then 1/x, 1/y, 1/z are in H.P. (Harmonic Progression).</p><p><strong>Step 3:</strong> For H.P., we know: 2/y ≠ 1/x + 1/z in general. Instead: 1/x, 1/y, 1/z satisfy 2/y = 1/x + 1/z only when x = z.</p><p><strong>Step 4:</strong> A key result: If x, y, z are positive and in A.P., then:</p><p>• xy + yz > xz (always true)</p><p>• x + z > 2√(xz) is false (equality holds when x = z)</p><p>• The most common correct statement: <strong>y² < xz is FALSE</strong> (actually y² > xz when x ≠ z)</p><p>• Correct: <strong>x/z + z/x > 2 when x ≠ z</strong> (by AM-GM)</p><p>∴ Without seeing options, typical answer involves: <strong>xy + yz > xz</strong> or similar product/quotient inequality</p>
Correct Answer: C