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

\frac{1}{2}

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\frac{1}{4}

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\frac{m + n}{6 m n}

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Solution

The correct option is B $\frac{1}{4}$
$\frac{x^{m} y^{n}}{(1 + x^{2 m}) (1 + y^{2 n)}}$

Divide by $x^{m} y^{n}$

$= \frac{1}{(\frac{1}{x^{m}} + x^{m}) (\frac{1}{y^{n}} + y^{n})}$

by $A . M \geq G . M$

$\frac{1}{x^{m}} + x^{m} \geq 2$ , $\frac{1}{y^{n}} + y^{n} \geq 2$

There by maximum value of $\frac{x^{m} y^{n}}{(1 + x^{2 m}) (1 + y^{2 n})} = \frac{1}{2 \times 2} = \frac{1}{4}$

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