Let $x, y$ be positive real numbers and $m, n$ positive integers. The maximum value of the expression…
Let $x, y$ be positive real numbers and $m, n$ 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
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{m+n}{6 m n}$
Solution
$A=\frac{x^{m} y^{n}}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)}=\frac{1}{\left(x^{-m}+x^{m}\right)\left(y^{-n}+y^{n}\right)}$
$\frac{x^{m}+y^{-m}}{2} \geq\left(x^{n+1}, x^{-m}\right)^{\frac{1}{2}} \Rightarrow x^{n+}+x^{-n} \geq 2$
In the same way, $y^{n}+y^{n} \geq 2$
Then $,\left(x^{n}+x^{-n}\right)\left(y^{-n}+y^{n}\right) \geq 4$
$\Rightarrow \frac{1}{\left(x^{\text {"1 }}+x^{-m}\right)\left(y^{-n}+y^{n}\right)} \leq \frac{1}{4}$