Question

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

• A. $\frac{1}{4}$
• B. $1$
• C. $\frac{m+n}{6mn}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{4}$

Let $T=\frac{x^m{y}^n}{(1+x^{2m})(1+y^{2n})}$

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

Using AM-GM inequality,
$x^{-m}+x^m\ge 2$
and $y^{-n}+y^n\ge 2$
$\therefore (x^{-m}+x^m)(y^{-m}+y^m)\ge 4$
$\frac{1}{(x^{-m}+x^m)(y^{-m}+y^m)}\le \frac{1}{4}$

At $x=0$, $T=\frac{1}{4}$
$\therefore$ Maximum value $=\frac{1}{4}$