Sequences And Series Ques 31
- Let $x, y$ be positive real numbers and $m, n$ positive integers. The maximum value of the expression $\frac{x^m y^n}{\left(1+x^{2 m}\right)\left(1+y^{2 n}\right)}$ is
(2019 Main, 11 Jan II)
(a) $\frac{1}{2}$
(b) $1$
(c) $\frac{1}{4}$
(d) $\frac{m+n}{6 m n}$
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Answer:
Correct Answer: 31.(c)
Solution: (c) Consider, $\frac{x^m y^n}{\left(1+x^{2 m}\right)\left(1+y^{2 m}\right)}$
$ =\frac{1}{\left(x^m+x^{-m}\right)\left(y^n+y^{-n}\right)} $
By using AM $\geq \mathrm{GM}$
(because $x, y \in R^{+}$), we get
$\left(x^m+x^{-m}\right) \geq 2$ and $\left(y^n+y^{-n}\right) \geq 2$
$\left[\because\right.$ If $x>0$, then $\left.x+\frac{1}{x} \geq 2\right]$
$\Rightarrow\left(x^m+x^{-m}\right)\left(y^n+y^{-n}\right) \geq 4$
$\Rightarrow \frac{1}{\left(x^m+x^{-m}\right)\left(y^n+y^{-n}\right)} \leq \frac{1}{4}$
$\therefore$ Maximum value $=\frac{1}{4}$.
BETA
