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  • Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression <img alt="\frac{x^{m}y^{n}}{(1+x^{2m})(1+y^{2n})}" src="https://entrancecorner.oncodecogs.com/

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

Option: 1

1


Option: 2

0.5


Option: 3

0.25


Option: 4

1.5


Answers (1)

best_answer

Relation between AM, GM and HM of two positive numbers -

AM\geqslant GM\geqslant HM

 

Now,

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

A.M.\geq G.M

\frac{\left (x^{m}+\frac{1}{x^{m}} \right )}{2}\geq\sqrt{\left (x^{m}\cdot \frac{1}{x^{m}} \right)}\quad\text{and}\quad\frac{\left (y^{m}+\frac{1}{y^{m}} \right )}{2}\geq\sqrt{\left (y^{m}\cdot \frac{1}{y^{m}} \right)}

{\left (x^{m}+\frac{1}{x^{m}} \right )}\geq2\quad\text{and}\quad{\left (y^{m}+\frac{1}{y^{m}} \right )}\geq2

Multiplying both

{\left (x^{m}+\frac{1}{x^{m}} \right )}\cdot{\left (y^{m}+\frac{1}{y^{m}} \right )}\geq4

So, \frac{1}{(x^{m}+\frac{1}{x^{m}})(y^{n}+\frac{1}{y^{n}})}=\frac{x^{m}y^{n}}{(1+x^{2m})(1+y^{2n})}\leq \frac{1}{4}

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