Let $x, y$ be positive real numbers and $\mathrm{m}, \mathrm{n}$ positive integers.
The maximum value of the expression $\frac{x^{\mathrm{m}} y^{\mathrm{n}}}{\left(1+x^{2 \mathrm{~m}}\right)\left(1+y^{2 \mathrm{n}}\right)}$
is:
Correct Option: , 3
$A=\frac{x^{m} y^{n}}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)}=\frac{1}{\left(x^{-m}+x^{m}\right)\left(y^{-n}+y^{n}\right)}$
$\frac{x^{m}+y^{-m}}{2} \geq\left(x^{m} \cdot x^{-m}\right)^{\frac{1}{2}} \Rightarrow x^{m}+x^{-m} \geq 2$
In the same way, $y^{-n}+y^{n} \geq 2$
Then, $\left(x^{m}+x^{-m}\right)\left(y^{-n}+y^{n}\right) \geq 4$
$\Rightarrow \quad \frac{1}{\left(x^{m}+x^{-m}\right)\left(y^{-n}+y^{n}\right)} \leq \frac{1}{4}$
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