Question

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

• A. 1/2
• B. 1/4
• C. m/n
• D. n/m

Answer 1

Answer: (B)

1/4

Answer 2

The expression is $E = \frac{x^my^n}{(1+x^{2n})(1+y^{2n})}$. This can be written as $E = \left(\frac{x^m}{1+x^{2n}}\right) \left(\frac{y^n}{1+y^{2n}}\right)$.

Let $f(t) = \frac{t^k}{1+t^{2l}}$. For $f(t)$ to have a finite maximum, the power in the numerator ($k$) must be less than twice the power in the denominator ($2l$). That is, $k < 2l$.

If $k < 2l$, the maximum occurs when $t^{2l} = \frac{k}{2l-k}$.

For the term $\frac{y^n}{1+y^{2n}}$, we have $k=n$ and $l=n$. The condition $n < 2n$ is always true for positive integer $n$. The maximum occurs when $y^{2n} = \frac{n}{2n-n} = 1$, which means $y=1$. The maximum value of this term is $\frac{1^n}{1+1^{2n}} = \frac{1}{2}$.

For the term $\frac{x^m}{1+x^{2n}}$, we have $k=m$ and $l=n$. For a finite maximum to exist, we must have $m < 2n$. If this condition is met, the maximum of this term is $\frac{2n-m}{2n} \left(\frac{m}{2n-m}\right)^{m/(2n)}$.

Since the problem asks for "the maximum value" (a single numerical value), it implies that this value is constant regardless of $m, n$ or that it's for a specific, common scenario. The simplest scenario that yields a constant maximum is when $m=n$.

If $m=n$, then the term $\frac{x^n}{1+x^{2n}}$ also has its maximum value at $x=1$, which is $\frac{1}{2}$.

In this case ($m=n$), the maximum value of the expression $E$ is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.