Let x, y be positive real numbers and m n positive integers.
Question:

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:

1. (1) 1

2. (2) $\frac{1}{2}$

3. (3) $\frac{1}{4}$

4. (4) $\frac{\mathrm{m}+\mathrm{n}}{6 \mathrm{mn}}$

Correct Option: , 3

Solution:

$A=\frac{x^{m} y^{n}}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)}=\frac{1}{\left(x^{-m}+x^{m}\right)\left(y^{-n}+y^{n}\right)}$

$\frac{x^{m}+y^{-m}}{2} \geq\left(x^{m} \cdot x^{-m}\right)^{\frac{1}{2}} \Rightarrow x^{m}+x^{-m} \geq 2$

In the same way, $y^{-n}+y^{n} \geq 2$

Then, $\left(x^{m}+x^{-m}\right)\left(y^{-n}+y^{n}\right) \geq 4$

$\Rightarrow \quad \frac{1}{\left(x^{m}+x^{-m}\right)\left(y^{-n}+y^{n}\right)} \leq \frac{1}{4}$