Let x, y be positive real numbers and m, n be | vedclass

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(C) We are given the expression $E = \frac{x^m y^n}{(1 + x^{2m})(1 + y^{2n})}$.We can rewrite this expression as $E = \frac{x^m}{1 + x^{2m}} 	imes \f

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$\frac{1}{4}$

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(C) We are given the expression $E = \frac{x^m y^n}{(1 + x^{2m})(1 + y^{2n})}$.We can rewrite this expression as $E = \frac{x^m}{1 + x^{2m}} 	imes \frac{y^n}{1 + y^{2n}}$.Dividing the numerator and denominator of each fraction by $x^m$ and $y^n$ respectively,we get:$E = \frac{1}{\left( \frac{1}{x^m} + x^m ight)} 	imes \frac{1}{\left( \frac{1}{y^n} + y^n ight)}$.By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for any positive real number $a$,$a + \frac{1}{a} \ge 2$,with equality holding when $a = 1$.Therefore,$x^m + \frac{1}{x^m} \ge 2$ and $y^n + \frac{1}{y^n} \ge 2$.This implies that $\frac{1}{x^m + \frac{1}{x^m}} \le \frac{1}{2}$ and $\frac{1}{y^n + \frac{1}{y^n}} \le \frac{1}{2}$.Multiplying these inequalities,we get the maximum value of $E$ as $\frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$.

Let $x, y$ be positive real numbers and $m, n$ be positive integers. The maximum value of the expression $\frac{x^m y^n}{(1 + x^{2m})(1 + y^{2n})}$ is

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