Question

If x and y are positive real numbers and m,n are any positive integers, then
$\frac{x^n.y^m}{(1+x^{2n})(1+y^{2m})}$ $>$$\frac{1}{4}$

• A. True
• B. False

Answer 1

The correct option is B

False

Given $\frac{x^n.y^m}{(1+x^{2n})(1+y^{2m})}$ $>$ $\frac{1}{4}$

$\Rightarrow$ $(1+x^{2n})(1+y^{2m})$ < 4 $x^n{y}^m$ ----------------(1)

Consider 1 , $x^{2n}$

As AM $\ge$ GM

$\frac{1+x^{2n}}{2}\ge 2\sqrt{x^{2n}}$

$\Rightarrow$ 1 + $x^{2n}$ $\ge$ 2$x^n$ ---------------------------(2)

Similarly for 1, $y^{2n}$

As AM $\ge$ GM

$\frac{1+y^{2n}}{2}\ge 2\sqrt{y^{2n}}$

$\Rightarrow$ 1 + $y^{2n}$ $\ge$ 2$y^n$ ---------------------------(3)

Multiplying (2) and (3)

$(1+x^{2n})(1+y^{2m})\ge 4x^n{y}^m$ Which is in contradiction with the given statement.

So, given statement is false.