<p>Let <em>x</em>, <em>y</em> be positive real numbers and <em>m</em>, <em>n</em> positive integers. The maximum value of the expression \(\dfrac{x^m y^n}{(1+x^{2m})(1+y^{2n})}\) is:</p>
Step-by-Step Solution
Key Concept: Apply AM-GM inequality to the denominators by recognizing that for positive terms, 1 + a² ≥ 2a, with equality when a = 1. This transforms the optimization problem into finding when the denominator is minimized relative to the numerator.
<p><strong>Step 1:</strong> Apply AM-GM to each factor in the denominator.</p><p>For the first term: 1 + x^(2m) ≥ 2√(x^(2m)) = 2x^m, with equality when x^(2m) = 1, i.e., x^m = 1</p><p>For the second term: 1 + y^(2n) ≥ 2√(y^(2n)) = 2y^n, with equality when y^(2n) = 1, i.e., y^n = 1</p><p><strong>Step 2:</strong> Substitute into the expression:</p><p>$$\frac{x^m y^n}{(1+x^{2m})(1+y^{2n})} \leq \frac{x^m y^n}{(2x^m)(2y^n)} = \frac{x^m y^n}{4x^m y^n} = \frac{1}{4}$$</p><p><strong>Step 3:</strong> Verify equality condition. Maximum is achieved when x^m = 1 and y^n = 1 (i.e., x = 1 and y = 1), giving value 1/4.</p><p>∴ Answer: <strong>1/4</strong> (Option C)</p>
Correct Answer: C