Question:
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression \(\frac{x^m y^n}{(1 + x^{2m})(1 + y^{2n})}\) is :
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression \(\frac{x^m y^n}{(1 + x^{2m})(1 + y^{2n})}\) is :
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Split the expression into x^m/(1+x^{2m}) times y^n/(1+y^{2n}) and bound each factor by AM-GM.
Updated On: Nov 5, 2026
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)
- \(\frac{m+n}{6mn}\)
- 1
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The Correct Option is B
Solution and Explanation
The expression equals \(\frac{x^m}{1+x^{2m}}\cdot\frac{y^n}{1+y^{2n}}\).
For positive x, \(1 + x^{2m} \ge 2x^m\) (AM-GM), so \(\frac{x^m}{1+x^{2m}} \le \frac12\), with equality at x = 1. Similarly \(\frac{y^n}{1+y^{2n}} \le \frac12\), with equality at y = 1.
Hence the expression is at most \(\frac12\cdot\frac12 = \frac14\), attained at x = y = 1.
Answer: (2) \(\frac{1}{4}\)
For positive x, \(1 + x^{2m} \ge 2x^m\) (AM-GM), so \(\frac{x^m}{1+x^{2m}} \le \frac12\), with equality at x = 1. Similarly \(\frac{y^n}{1+y^{2n}} \le \frac12\), with equality at y = 1.
Hence the expression is at most \(\frac12\cdot\frac12 = \frac14\), attained at x = y = 1.
Answer: (2) \(\frac{1}{4}\)
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