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:

We have to maximise the real-valued expression

$$E=\dfrac{x^{m}y^{n}}{\left(1+x^{2m}\right)\left(1+y^{2n}\right)}$$

where $$x>0,\;y>0$$ and $$m,n$$ are positive integers.

Because the variables $$x$$ and $$y$$ occur in separate factors of the denominator, we factorise the expression as a product of two independent functions:

$$E=\left(\dfrac{x^{m}}{1+x^{2m}}\right)\left(\dfrac{y^{n}}{1+y^{2n}}\right).$$

Let us denote

$$f(x)=\dfrac{x^{m}}{1+x^{2m}},\qquad g(y)=\dfrac{y^{n}}{1+y^{2n}}.$$

Clearly $$E=f(x)\,g(y).$$ To maximise $$E$$ we can first find the individual maxima of $$f(x)$$ and $$g(y),$$ because the two functions depend on different variables and both attain their maxima at positive points.

We concentrate on $$f(x).$$ Set

$$a=x^{m}\quad(a>0).$$

Then

$$f(x)=\dfrac{a}{1+a^{2}}.$$

To obtain an upper bound for $$\dfrac{a}{1+a^{2}},$$ we invoke the AM-GM inequality.

For any positive $$a$$ we have

$$1+a^{2}\ge 2\sqrt{1\cdot a^{2}}=2a.$$

Dividing both sides by the positive quantity $$1+a^{2}$$ gives

$$\dfrac{a}{1+a^{2}}\le\dfrac{a}{2a}=\dfrac12.$$

Equality in the AM-GM step occurs precisely when the two terms inside the mean are equal, that is, when

$$1=a^{2}\;\Longrightarrow\;a=1.$$

Because $$a=x^{m},$$ this equality condition translates to

$$x^{m}=1\;\Longrightarrow\;x=1.$$

Therefore the maximum value of $$f(x)$$ is

$$f_{\max}=\dfrac12,\quad\text{attained at }x=1.$$

Exactly the same reasoning applies to $$g(y).$$ Setting $$b=y^{n}\;(b>0)$$ we obtain

$$g(y)=\dfrac{b}{1+b^{2}}\le\dfrac12,$$

with equality when $$b=1,$$ i.e. when $$y^{n}=1\;\Longrightarrow\;y=1.$$

Hence

$$g_{\max}=\dfrac12.$$

Finally, substituting the individual maxima into the product $$E=f(x)\,g(y)$$ we get

$$E_{\max}=f_{\max}\,g_{\max}=\dfrac12\cdot\dfrac12=\dfrac14.$$

This upper bound is actually obtainable, because the equalities in both AM-GM applications can hold simultaneously at

$$x=1,\;y=1.$$

So the maximum of the given expression is $$\dfrac14.$$ Hence, the correct answer is Option C.